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(Hindi Version of Business Statistics)
BUSINESS STATISTICS
For B.Com. (Pass and Honours) ; B.A. (Economics Honours) ;
M.B.A./M.M.S. of Indian Universities
S.C. GUPTA
M.A. (Statistics) ; M.A. (Mathematics) ; M.S. (U.S.A.)
Associate Professor in Statistics (Retired)
Hindu College, University of Delhi
Delhi-110007
Mrs. INDRA GUPTA
(iv)
©
Author
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or by any means, electronic, mechanical, photocopying, recording and/or otherwise without the prior
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(v)
Dedicated to
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(vii)
Preface
TO THE SIXTH EDITION
The book originally written over 20 years ago has been revised and reprinted several times during the
intervening period. It is very heartening to note that there has been an increasing response for the book from
the students of B.A. (Economics Honours), B.Com. (Pass and Honours); M.B.A./M.M.S. and other
management courses, in spite of the fact that the book has not been revised for quite a long time. I take
great pleasure in presenting to the readers, the sixth thoroughly revised and enlarged edition of the book.
The book has been revised in the light of the valuable criticism, suggestions and the feedback received from
the teachers, students and other readers of the book from all over the country.
Some salient features of the new edition are :
•
The theoretical discussion throughout has been refined, restructured, rewritten and updated.
During the course of rewriting, a sincere attempt has been made to retain the basic features of the
earlier editions viz., the simplicity of presentation, lucidity of style and the analytical approach,
which have been appreciated by the teachers and the students all over India.
•
Several new topics have been added at appropriate places to make the treatment of the subject
matter more exhaustive and up-to-date. Some of the additions are given below :
Remark
Remark 4
Remark 6
•
,
,
,
page 5·57
page 6·3
page 6·10
: Effect of Change of Scale on Harmonic Mean.
: Effect of Change of Origin and Scale on Range.
: Effect of Change of Origin and Scale on Mean Deviation about
Mean.
: Xmax and X min in terms of Mean and Range.
: Some Results on Covariance.
: Lag and Lead Correlation.
: Necessary and Sufficient Condition for Minima of E.
Remark 6
, page 7·3
Remark 2
, page 8·12
§ 8·10
, page 8·45
Remark 1 ⎤ , page 9·4
⎥
and
⎥
Theorem ⎥⎦
Remark
, page 9·24 : Limits for r.
§ 11·9
, page 11·54 : Time Series Analysis in Forecasting.
§ 13·9
, page 13·9 : Covariance In Terms of Expectation.
§ 13·10
, page 13·14 : Var (ax + by) = a2 Var (x) + b2 Var (y) + 2ab Cov (x, y)
and Remark
Equations (14·29e)
—
and (14·29f) , page 14·31 : Distribution of the Mean (X ) of n i.i.d. N (μ, σ2) variates.
§ 15·11·2
, page 15·18 : Sampling Distribution of Mean.
– 15·19
A number of solved examples, selected from the latest examination papers of various universities
and professional institutes, have been added. These are bound to assist understanding and provide
greater variety.
(viii)
•
Exercise sets containing questions and unsolved problems at the end of each Chapter have been
substantially reorganised and rewritten by deleting old problems and adding new problems,
selected from the latest examination papers of various universities, C.A., I.C.W.A., and other
management courses. All the problems have been very carefully graded and answers to the
problems are given at the end of each problem.
•
An attempt has been made to rectify the errors in the last edition.
It is hoped that all these changes, additions and improvements will enhance the value of the book. We
are confident, that the book in its present form, will prove to be of much greater utility to the students as
well as teachers of the subject.
We express our deep sense of thanks and gratitude to our publishers M/s Himalaya Publishing House
and Type-setters, M/s Times Printographics, Darya Ganj, New Delhi, for their untiring efforts, unfailing
courtesy, and co-operation in bringing out the book in such an elegant form.
We strongly believe that the road to improvement is never-ending. Suggestions and criticism for
further improvement of the book will be very much appreciated and most gratefully acknowledged.
January, 2013
S.C. GUPTA
INDRA GUPTA
(ix)
Preface
TO THE FIRST EDITION
In the ancient times Statistics was regarded only as the science of statecraft and was used to collect
information relating to crimes, military strength, population, wealth, etc., for devising military and fiscal
policies. But today, Statistics is not merely a by-product of the administrative set-up of the State but it
embraces all sciences-social, physical and natural, and is finding numerous applications in various
diversified fields such as agriculture, industry, sociology, biometry, planning, economics, business,
management, insurance, accountancy and auditing, and so on. Statistics (theory and methods) is used
extensively by the government or business or management organisations in planning future programmes
and formulating policy decisions. It is rather impossible to think of any sphere of human activity where
Statistics does not creep in. The subject of Statistics has acquired tremendous progress in the recent past so
much so that an elementary knowledge of statistical methods has become a part of the general education in
the curricula of many academic and professional courses.
This book is a modest though determined bid to serve as a text-book, for B.Com. (Pass and Hons.);
B.A. Economics (Hons.) courses of Indian Universities. The main aim in writing this book is to present a
clear, simple, systematic and comprehensive exposition of the principles, methods and techniques of
Statistics in various disciplines with special reference to Economics and Business. The stress is on the
applications of techniques and methods most commonly used by statisticians. The lucidity of style and
simplicity of expression have been our twin objectives in preparing this text. Mathematical complexity has
been avoided as far as possible. Wherever desirable, the notations and terminology have been clearly
explained and then all the mathematical steps have been explained in detail.
An attempt has been made to start with the explanation of the elementaries of a topic and then the
complexities and the intricacies of the advanced problems have been explained and solved in a lucid
manner. A number of typical problems mostly from various university examination papers have been
solved as illustrations so as to expose the students to different techniques of tackling the problems and
enable them to have a better and throughful understanding of the basic concepts of the theory and its
various applications. At many places explanatory remarks have been given to widen readers’ horizon.
Moreover, in order to enable the readers to have a proper appreciation of the subject-matter and to fortify
their confidence in the understanding and application of methods, a large number of carefully graded
problems, mostly drawn from various university examination papers, have been given as exercise sets in
each chapter. Answers to all the problems in the exercise sets are given at the end of each problem.
The book contains 16 Chapters. We will not enumerate the topics discussed in the text since an idea of
these can be obtained from a cursory glance at the table of contents. Chapters 1 to 11 are devoted to
‘Descriptive Statistics’ which consists in describing some characteristics like averages, dispersion,
skewness, kurtosis, correlation, etc., of the numerical data. In spite of many latest developments in
statistical techniques, the old topics like ‘Classification and Tabulation’ (Chapter 3) and ‘Diagrammatic and
Graphic Representation’ (Chapter 4) have been discussed in details, since they still constitute the bulk of
statistical work in government and business organisations. The use of statistical methods as scientific tools
in the analysis of economic and business data has been explained in Chapter 10 (Index Numbers) and
Chapter 11 (Times Series Analysis). Chapters 12 to 14 relate to advanced topics like Probability, Random
(x)
Variable, Mathematical Expectation and Theoretical Distributions. An attempt has been made to give a
detailed discussion of these topics on modern lines through the concepts of ‘Sample Space’ and ‘Axiomatic
Approach’ in a very simple and lucid manner. Chapter 15 (Sampling and Design of Sample Surveys),
explains the various techniques of planning and executing statistical enquiries so as to arrive at valid
conclusions about the population. Chapter 16 (Interpolation and Extrapolation) deals with the techniques of
estimating the value of a function y = f (x) for any given intermediate value of the variable x.
We must unreservedly acknowledge our deep debt of gratitude we owe to the numerous authors whose
great and masterly works we have consulted during the preparation of the manuscript.
We take this opportunity to express our sincere gratitude to Prof. Kanwar Sen, Shri V.K. Kapoor and a
number of students for their valuable help and suggestions in the preparation of this book.
Last but not least, we express our deep sense of gratitude to our Publishers M/s Himalaya Publishing
House for their untiring efforts and unfailing courtesy and co-operation in bringing out the book in time in
such an elegant form.
Every effort has been made to avoid printing errors though some might have crept in inadvertently. We
shall be obliged if any such errors are brought to our notice. Valuable suggestions and criticism for the
improvement of the book from our colleagues (who are teaching this course) and students will be highly
appreciated and duly incorporated in subsequent editions.
June, 1988
S.C. GUPTA
Mrs. INDRA GUPTA
(xi)
Contents
1.
INTRODUCTION — MEANING AND SCOPE
1·1 – 1·16
1·1.
1·2.
1·3.
1·4.
1·5.
ORIGIN AND DEVELOPMENT OF STATISTICS
DEFINITION OF STATISTICS
IMPORTANCE AND SCOPE OF STATISTICS
LIMITATIONS OF STATISTICS
DISTRUST OF STATISTICS
EXERCISE 1.1 .
1·1
1·2
1·5
1·11
1·12
1·13
2.
COLLECTION OF DATA
2·1.
2·1·1.
2·1·2.
2·1·3.
2·1·4.
2·1·5.
2·1·6.
2·2.
2·2·1.
2·3.
2·3·1.
2·3·2.
2·3·3.
2·3·4.
2·3·5.
2·4.
2·5.
2·5·1.
2·5·2.
2·6.
INTRODUCTION
Objectives and Scope of the Enquiry.
Statistical Units to be Used.
Sources of Information (Data).
Methods of Data Collection.
Degree of Accuracy Aimed at in the Final Results.
Type of Enquiry.
PRIMARY AND SECONDARY DATA
Choice Between Primary and Secondary Data.
METHODS OF COLLECTING PRIMARY DATA
Direct Personal Investigation.
Indirect Oral Investigation.
Information Received Through Local Agencies.
Mailed Questionnaire Method.
Schedules Sent Through Enumerators.
DRAFTING OR FRAMING THE QUESTIONNAIRE
SOURCES OF SECONDARY DATA
Published Sources.
Unpublished Sources.
PRECAUTIONS IN THE USE OF SECONDARY DATA
EXERCISE 2·1.
3.
CLASSIFICATION AND TABULATION
3·1.
3·2.
3·2·1.
3·2·2.
INTRODUCTION – ORGANISATION OF DATA
CLASSIFICATION
Functions of Classification.
Rules for Classification.
2·1 – 2·21
2·1
2·1
2·2
2·4
2·4
2·4
2·5
2·6
2·7
2·7
2·8
2·8
2·9
2·10
2·11
2·12
2·16
2·16
2·18
2·18
2·20
3·1 – 3·40
3·1
3·1
3·2
3·2
(xii)
3·2·3.
3·3.
3·3·1.
3·3·2.
3·3·3.
3·3·4.
3·4.
3·4·1.
3·4·2.
3·4·3.
3·4·4.
3·5.
3·5·1.
3·5·2.
3·6.
3·7.
3·7·1.
3·7·2.
3·7·3.
Bases of Classification.
FREQUENCY DISTRIBUTION
Array
Discrete or Ungrouped Frequency Distribution.
Grouped Frequency Distribution.
Continuous Frequency Distribution.
BASIC PRINCIPLES FOR FORMING A GROUPED FREQUENCY DISTRIBUTION
Types of Classes.
Number of Classes.
Size of Class Intervals.
Types of Class Intervals.
CUMULATIVE FREQUENCY DISTRIBUTION
Less Than Cumulative Frequency.
More Than Cumulative Frequency.
BIVARIATE FREQUENCY DISTRIBUTION
EXERCISE 3·1 .
TABULATION – MEANING AND IMPORTANCE
Parts of a Table.
Requisites of a Good Table.
Types of Tabulation.
EXERCISE 3·2 .
4.
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
4·1.
4·2.
4·3.
4·3·1.
4·3·2.
4·3·3.
4·3·4.
4·3·5.
4·3·6.
4·3·7.
4·3·8.
INTRODUCTION
DIFFERENCE BETWEEN DIAGRAMS AND GRAPHS
DIAGRAMMATIC PRESENTATION
General Rules for Constructing Diagrams.
Types of Diagrams.
One-dimensional Diagrams.
Two-dimensional Diagrams.
Three-Dimensional Diagrams.
Pictograms
Cartograms
Choice of a Diagram.
EXERCISE 4·1
4·4.
4·4·1.
4·4·2.
4·4·3.
4·4·4.
4·4·5.
4·5.
GRAPHIC REPRESENTATION OF DATA
Technique of Construction of Graphs.
General Rules for Graphing.
Graphs of Frequency Distributions.
Graphs of Time Series or Historigrams.
Semi-Logarithmic Line Graphs or Ratio Charts.
LIMITATIONS OF DIAGRAMS AND GRAPHS
EXERCISE 4·2
3·3
3·6
3·6
3·6
3·7
3·8
3·8
3·8
3·8
3·10
3·11
3·17
3·18
3·18
3·20
3·22
3·27
3·27
3·29
3·30
3·38
4·1 – 4·57
4·1
4·1
4·2
4·2
4·3
4·3
4·12
4·20
4·22
4·24
4·24
4·24
4·27
4·27
4·28
4·29
4·40
4·47
4·53
4·53
(xiii)
5.
AVERAGES OR MEASURES OF CENTRAL TENDENCY
5·1 – 5·68
5·1.
5·2.
5·3.
5·4.
5·4·1.
5·4·2.
5·4·3.
5·5.
INTRODUCTION
5·1
REQUISITES OF A GOOD AVERAGE OR MEASURE OF CENTRAL TENDENCY 5·2
VARIOUS MEASURES OF CENTRAL TENDENCY
5·2
ARITHMETIC MEAN
5·2
Step Deviation Method for Computing Arithmetic Mean.
5·3
Mathematical Properties of Arithmetic Mean.
5·5
Merits and Demerits of Arithmetic Mean.
5·8
WEIGHTED ARITHMETIC MEAN
5·14
EXERCISE 5·1
5·17
5·6.
MEDIAN
5·22
5·6·1. Calculation of Median.
5·22
5·6·2. Merits and Demerits of Median.
5·24
5·6·3. Partition Values.
5·26
5·6·4. Graphic Method of Locating Partition Values.
5·28
EXERCISE 5·2
5·31
5·7.
MODE
5·35
5·7·1. Computation of Mode.
5·36
5·7·2. Merits and Demerits of Mode.
5·37
5·7·3. Graphic Location of Mode.
5·38
5·8.
EMPIRICAL RELATION BETWEEN MEAN (M), MEDIAN (Md) AND MODE (Mo) 5·38
EXERCISE 5·3
5·45
5·9.
GEOMETRIC MEAN
5·49
5·9·1. Merits and Demerits of Geometric Mean.
5·50
5·9·2. Compound Interest Formula.
5·51
5·9·3. Average Rate of a Variable Which Increases by Different Rates at Different Periods.
5·51
5·9·4. Wrong Observations and Geometric Mean.
5·52
5·9·5. Weighted Geometric Mean.
5·56
5·10. HARMONIC MEAN
5·57
5·10·1. Merits and Demerits of Harmonic Mean.
5·57
5·10·2. Weighted Harmonic Mean.
5·61
5·11. RELATION BETWEEN ARITHMETIC MEAN, GEOMETRIC MEAN
AND HARMONIC MEAN
5·61
5·12. SELECTION OF AN AVERAGE
5·62
5·13. LIMITATIONS OF AVERAGES
5·63
EXERCISE 5·4
5·63
6.
6·1.
6·1·1.
6·2.
6·3.
6·4.
6·5.
MEASURES OF DISPERSION
INTRODUCTION AND MEANING
Objectives or Significance of the Measures of Dispersion.
CHARACTERISTICS FOR AN IDEAL MEASURE OF DISPERSION
ABSOLUTE AND RELATIVE MEASURES OF DISPERSION
MEASURES OF DISPERSION
RANGE
6·1 – 6·53
6·1
6·2
6·2
6·2
6·2
6·3
(xiv)
6·5·1.
6·5·2.
6·6.
6·6·1.
6·7.
Merits and Demerits of Range.
Uses.
QUARTILE DEVIATION OR SEMI INTER-QUARTILE RANGE
Merits and Demerits of Quartile Deviation.
PERCENTILE RANGE
EXERCISE 6·1
6·8.
6·8·1.
6·8·2.
6·8·3.
6·8·4.
6·8·5.
MEAN DEVIATION OR AVERAGE DEVIATION
Computation of Mean Deviation.
Short-cut Method of Computing Mean Deviation.
Merits and Demerits of Mean Deviation.
Uses.
Relative Measure of Mean Deviation.
EXERCISE 6·2
6·9.
6·9·1.
6·9·2.
6·9·3.
6·9·4.
STANDARD DEVIATION
Mathematical Properties of Standard Deviation.
Merits and Demerits of Standard Deviation
Variance and Mean Square Deviation.
Different Formulae for Calculating Variance.
EXERCISE 6·3
6·10.
6·11.
6·12.
STANDARD DEVIATION OF THE COMBINED SERIES
COEFFICIENT OF VARIATION
RELATIONS BETWEEN VARIOUS MEASURES OF DISPERSION
EXERCISE 6·4
6·13.
LORENZ CURVE
EXERCISE 6·5
EXERCISE 6·6
7.
SKEWNESS, MOMENTS AND KURTOSIS
7·1.
7·2.
7·2·1.
7·2·2.
INTRODUCTION
SKEWNESS
Measures of Skewness.
Karl Pearson’s Coefficient of Skewness.
EXERCISE 7·1
7·2·3.
7·2·4.
7·2·5.
Bowley’s Coefficient of Skewness.
Kelly’s Measure of Skewness.
Coefficient of Skewness based on Moments.
EXERCISE 7·2
7·3.
7·3·1.
7·3·2.
7·3·3.
7·3·4.
7·3·5.
MOMENTS
Moments about Mean.
Moments about Arbitrary Point A.
Relation between Moments about Mean and Moments about Arbitrary Point ‘A’.
Effect of Change of Origin and Scale on Moments about Mean.
Sheppard’s Correction for Moments.
6·4
6·4
6·5
6·5
6·6
6·7
6·9
6·9
6·10
6·11
6·11
6·11
6·15
6·16
6·18
6·18
6·19
6·19
6·29
6·33
6·36
6·41
6·42
6·47
6·49
6·50
7·1 – 7·34
7·1
7·1
7·2
7·2
7·9
7·12
7·13
7·13
7·16
7·18
7·19
7·19
7·19
7·21
7·21
(xv)
7·3·6.
7·4.
7·5.
7·6.
Charlier Checks.
KARL PEARSON’S BETA (β) AND GAMMA (γ) COEFFICIENTS
BASED ON MOMENTS
COEFFICIENT OF SKEWNESS BASED ON MOMENTS
KURTOSIS
EXERCISE 7·3.
8.
CORRELATION ANALYSIS
8·1.
8·1·1.
8·1·2.
8·2.
8·3.
INTRODUCTION
Types of Correlation
Correlation and Causation.
METHODS OF STUDYING CORRELATION
SCATTER DIAGRAM METHOD
7·21
7·22
7·22
7·23
7·30
8·1 – 8·46
8.10.
8·1
8·1
8·2
8·3
8·3
EXERCISE 8·1
8·5
KARL PEARSON’S COEFFICIENT OF CORRELATION (COVARIANCE METHOD) 8·7
Properties of Correlation Coefficient
8·11
Assumptions Underlying Karl Pearson’s Correlation Coefficient.
8·17
Interpretation of r.
8·18
PROBABLE ERROR
8·18
EXERCISE 8·2
8·20
CORRELATION IN BIVARIATE FREQUENCY TABLE
8·25
EXERCISE 8·3
8·30
RANK CORRELATION METHOD
8·31
Limits for ρ.
8·31
Computation of Rank Correlation Coefficient (ρ).
8·32
Remarks on Spearman’s Rank Correlation Coefficient
8·38
EXERCISE 8·4
8·39
METHOD OF CONCURRENT DEVIATIONS
8·41
EXERCISE 8·5
8·43
COEFFICIENT OF DETERMINATION
8·43
EXERCISE 8·6
8·44
LAG AND LEAD CORRELATION
8.45
9.
LINEAR REGRESSION ANALYSIS
9·1.
9·2.
9·3.
9·3·1.
9·3·2.
9·3·3.
9·4.
9·4·1.
INTRODUCTION
LINEAR AND NON-LINEAR REGRESSION
LINES OF REGRESSION
Derivation of Line of Regression of y on x.
Line of Regression of x on y.
Angle Between the Regression Lines.
COEFFICIENTS OF REGRESSION
Theorems on Regression Coefficients
8·4.
8·4·1.
8·4·2.
8·4·3.
8·5.
8·6.
8·7.
8·7·1.
8·7·2.
8·7·3.
8·8.
8·9.
EXERCISE 9·1
9·1 – 9·33
9·1
9·2
9·2
9·2
9·4
9·5
9·6
9·7
9·15
(xvi)
9·5.
9·6.
9·7.
9·8.
9·9.
—
—
TO FIND THE MEAN VALUES (X , Y ) FROM THE TWO LINES OF
REGRESSION
TO FIND THE REGRESSION COEFFICIENTS AND THE CORRELATION
COEFFICIENT FROM THE TWO LINES OF REGRESSION
STANDARD ERROR OF AN ESTIMATE
REGRESSION EQUATIONS FOR A BIVARIATE FREQUENCY TABLE
CORRELATION ANALYSIS vs. REGRESSION ANALYSIS
EXERCISE 9·2
EXERCISE 9·3
10.
INDEX NUMBERS
10·1.
10·2.
10·3.
10·4.
10·5.
10·5·1.
10·5·2.
10·5·3.
10·5·4.
INTRODUCTION
USES OF INDEX NUMBERS
TYPES OF INDEX NUMBERS
PROBLEMS IN THE CONSTRUCTION OF INDEX NUMBERS
METHODS OF CONSTRUCTING INDEX NUMBERS
Simple (Unweighted) Aggregate Method.
Weighted Aggregate Method.
Simple Average of Price Relatives.
Weighted Average of Price Relatives
EXERCISE 10·1
10·6.
10·6·1.
10·6·2.
10·6·3.
10·6·4.
TESTS OF CONSISTENCY OF INDEX NUMBER FORMULAE
Unit Test.
Time Reversal Test.
Factor Reversal Test.
Circular Test.
EXERCISE 10·2
10·7.
10·7·1.
10·7·2.
CHAIN INDICES OR CHAIN BASE INDEX NUMBERS
Uses of Chain Base Index Numbers.
Limitations of Chain Base Index Numbers.
EXERCISE 10·3
10·8.
10·8·1.
10·8·2.
10·8·3.
BASE SHIFTING, SPLICING AND DEFLATING OF INDEX NUMBERS
Base Shifting.
Splicing.
Deflating of Index Numbers.
EXERCISE 10·4
10·9.
10·9·1.
10·9·2.
10·9·3.
10·10.
COST OF LIVING INDEX NUMBER
Main Steps in the Construction of Cost of Living Index Numbers
Construction of Cost of Living Index Numbers.
Uses of Cost of Living Index Numbers.
LIMITATIONS OF INDEX NUMBERS
EXERCISE 10·5
9·20
9·20
9·23
9·26
9·28
9·29
9·32
10·1 – 10·65
10·1
10·1
10·3
10·3
10·7
10·7
10·8
10·15
10.17
10·20
10·26
10·26
10·26
10·27
10·28
10·33
10·35
10·36
10·36
10·38
10·40
10·40
10·41
10·45
10·48
10·51
10·52
10·53
10·53
10·59
10·60
(xvii)
11.
TIME SERIES ANALYSIS
11·1.
11·2.
11·2·1.
11·2·2.
11·2·3.
11·3.
11·4.
11·5.
11·5·1.
11·5·2.
11·5·3.
11·5·4.
11·5·5.
INTRODUCTION
COMPONENTS OF A TIME SERIES
Secular Trend.
Short-Term Variations.
Random or Irregular Variations.
ANALYSIS OF TIME SERIES
MATHEMATICAL MODELS FOR TIME SERIES
MEASUREMENT OF TREND
Graphic or Free Hand Curve Fitting Method.
Method of Semi-Averages.
Method of Curve Fitting by the Principle of Least Squares.
Conversion of Trend Equation.
Selection of the Type of Trend.
EXERCISE 11·1
11·5·6.
Method of Moving Averages.
EXERCISE 11·2
11·6.
11·6·1.
11·6·2.
11·6·3.
11·6·4.
11·6·5.
11·7.
11·8.
11.9.
MEASUREMENT OF SEASONAL VARIATIONS
Method of Simple Averages.
Ratio to Trend Method.
‘Ratio to Moving Average’ Method.
Method of Link Relatives.
Deseasonalisation of Data.
MEASUREMENT OF CYCLICAL VARIATIONS
MEASUREMENT OF IRREGULAR VARIATIONS
TIME SERIES ANALYSIS IN FORECASTING
EXERCISE 11·3
EXERCISE 11·4
12.
THEORY OF PROBABILITY
12·1.
12·2.
12·3.
12·4.
12·4·1.
12·4·2.
12·5.
12·6.
INTRODUCTION
SHORT HISTORY
TERMINOLOGY
MATHEMATICAL PRELIMINARIES
Set Theory.
Permutation and Combination.
MATHEMATICAL OR CLASSICAL OR ‘A PRIORI’ PROBABILITY
STATISTICAL OR EMPIRICAL PROBABILITY
EXERCISE 12·1
12·7.
12·8.
12·8·1.
12·8·2.
AXIOMATIC PROBABILITY
ADDITION THEOREM OF PROBABILITY
Addition Theorem of Probability for Mutually Exclusive Events.
Generalisation of Addition Theorem of Probability.
11·1 – 11·60
11·1
11·1
11·2
11·3
11·4
11·5
11·5
11·6
11·6
11·7
11·10
11·22
11·25
11·25
11·30
11·38
11·39
11·40
11·42
11·44
11·47
11·49
11·52
11·53
11.54
11·54
11·59
12·1 – 12·52
12·1
12·1
12·2
12·4
12·4
12·6
12·8
12·9
12·14
12·17
12·19
12·20
12·20
(xviii)
12·9.
12·9·1.
12·9·2.
THEOREM OF COMPOUND PROBABILITY OR
MULTIPLICATION THEOREM OF PROBABILITY
Generalisation of Multiplication Theorem of Probability.
Independent Events.
Multiplication Theorem for Independent Events.
12·21
12·22
12·22
12·22
12·35
12·41
12·43
12·43
12·49
EXERCISE 12·2
OBJECTIVE TYPE QUESTIONS
12·10.
INVERSE PROBABILITY
Bayes’s Theorem (Rule for the Inverse Probability)
EXERCISE 12·3
13.
13·1.
13·2.
13·3.
13·3·1.
13·4.
13·5.
RANDOM VARIABLE, PROBABILITY DISTRIBUTIONS
AND MATHEMATICAL EXPECTATION
RANDOM VARIABLE
PROBABILITY DISTRIBUTION OF A DISCRETE RANDOM VARIABLE
PROBABILITY DISTRIBUTION OF A CONTINUOUS RANDOM VARIABLE
Probability Density Function (p.d.f.) of Continuous random Variable
DISTRIBUTION FUNCTION OR CUMULATIVE PROBABILITY FUNCTION
MOMENTS
EXERCISE 13·1
13·6.
13·7.
13·8.
13.9.
13·10.
13·11.
MATHEMATICAL EXPECTATION
Physical Interpretation of E(X).
THEOREMS ON EXPECTATION
VARIANCE OF X IN TERMS OF EXPECTATION
COVARIANCE IN TERMS OF EXPECTATION
VARIANCE OF LINEAR COMBINATION
JOINT AND MARGINAL PROBABILITY DISTRIBUTIONS
EXERCISE 13·2
14.
THEORETICAL DISTRIBUTIONS
14·1.
14·2.
14·2·1.
14·2·2.
14·2·3.
14·2·4.
INTRODUCTION
BINOMIAL DISTRIBUTION
Probability Function of Binomial Distribution.
Constants of Bionomial Distribution
Mode of Binomial Distribution.
Fitting of Binomial Distribution.
EXERCISE 14·1
14·3.
14·3·1.
14·3·2.
14·3·3.
14·3·4.
13·1 – 13·19
POISSON DISTRIBUTION
(AS A LIMITING CASE OF BINOMIAL DISTRIBUTION)
Utility or Importance of Poisson Distribution.
Constants of Poisson Distribution
Mode of Poisson Distribution
Fitting of Poisson Distribution.
EXERCISE 14·2
13·1
13·2
13·2
13·2
13·3
13·4
13·6
13·7
13·7
13·8
13·9
13·9
13·14
13·14
13·16
14·1 – 14·52
14·1
14·1
14·2
14·3
14·5
14·10
14·11
14·16
14·18
14·18
14·19
14·23
14·25
(xix)
14·4.
14·4·1.
14·4·2.
14·4·3.
14·4·4.
14·4·5.
14·4·6.
14·4·7.
NORMAL DISTRIBUTION
Equation of Normal Probability Curve.
Standard Normal Distribution.
Relation between Binomial and Normal Distributions.
Relation between Poisson and Normal Distributions.
Properties of Normal Distribution.
Areas Under Standard Normal Probability Curve
Importance of Normal Distribution.
EXERCISE 14·3
15.
SAMPLING THEORY AND DESIGN OF SAMPLE SURVEYS
15·1.
15·2.
15·3.
15·4.
15·4·1.
15·4·2.
15·5.
15·5·1.
15·5·2.
15·5·3.
15·5·4.
15·5·5.
15·6.
15·7.
15·8.
15·9.
15·9·1.
15·9·2.
15·9·3.
15·10.
15·10·1.
15·10·2.
15·10·3.
15·11.
15·11·1.
15·11·2.
15·11·3.
15·12.
15·12·1.
15·12·2.
15·13.
15·13·1.
INTRODUCTION
UNIVERSE OR POPULATION
SAMPLING
PARAMETER AND STATISTIC
Sampling Distribution.
Standard Error.
PRINCIPLES OF SAMPLING
Law of Statistical Regularity.
Principle of Inertia of Large Numbers.
Principle of Persistence of Small Numbers.
Principle of Validity.
Principle of Optimisation.
CENSUS VERSUS SAMPLE ENUMERATION
LIMITATIONS OF SAMPLING
PRINCIPAL STEPS IN A SAMPLE SURVEY
ERRORS IN STATISTICS
Sampling and Non-Sampling Errors.
Biased and Unbiased Errors.
Measures of Statistical Errors (Absolute and Relative Errors).
TYPES OF SAMPLING
Purposive or Subjective or Judgment Sampling.
Probability Sampling.
Mixed Sampling.
SIMPLE RANDOM SAMPLING
Selection of a Simple Random Sample.
Sampling Distribution of Mean
Merits and Limitations of Simple Random Sampling
STRATIFIED RANDOM SAMPLING
Allocation of Sample Size in Stratified Sampling.
Merits and Demerits of Stratified Random Sampling.
SYSTEMATIC SAMPLING
Merits and Demerits
14·28
14·28
14·29
14·29
14·30
14·30
14·33
14·36
14·45
15·1 – 15·26
15·1
15·1
15·2
15·2
15·3
15·3
15·4
15·4
15·5
15·5
15·5
15·5
15·5
15·7
15·8
15·10
15·10
15·13
15·14
15·14
15·14
15·15
15·15
15·15
15·16
15·18
15·19
15·19
15·20
15·21
15·22
15·23
(xx)
15·14.
15·15.
15·16.
CLUSTER SAMPLING
MULTISTAGE SAMPLING
QUOTA SAMPLING
15·23
15·23
15·24
15·24
EXERCISE 15·1.
16.
INTERPOLATION AND EXTRAPOLATION
16·1.
16·1·1.
16·1·2.
16·2.
16·3.
16·4.
16·5.
16·6.
16·7.
16·8.
INTRODUCTION
Assumptions.
Uses of Interpolation.
METHODS OF INTERPOLATION
GRAPHIC METHOD
ALGEBRAIC METHOD
METHOD OF PARABOLIC CURVE FITTING
METHOD OF FINITE DIFFERENCES
NEWTON’S FORWARD DIFFERENCE FORMULA
NEWTON’S BACKWARD DIFFERENCE FORMULA
16·1 – 16·28
EXERCISE 16·1
16·9.
BINOMIAL EXPANSION METHOD FOR INTERPOLATING MISSING VALUES
EXERCISE 16·2
16·10.
16·11.
16·11·1.
16·12.
16·13.
INTERPOLATION WITH ARGUMENTS AT UNEQUAL INTERVALS
DIVIDED DIFFERENCES
Newton’s Divided Difference Formula.
LAGRANGE’S FORMULA
INVERSE INTERPOLATION
EXERCISE 16·3
17.
INTERPRETATION OF DATA AND STATISTICAL FALLACIES
17·1.
17·2.
INTRODUCTION
INTERPRETATION OF DATA AND STATISTICAL FALLACIES –
MEANING AND NEED
FACTORS LEADING TO MIS-INTERPRETATION OF DATA
OR STATISTICAL FALLACIES
Bias.
Inconsistencies in Definitions.
Faulty Generalisations.
Inappropriate Comparisons.
Wrong Interpretation of Statistical Measures.
(a) Wrong Interpretation of Index Numbers.
(b) Wrong Interpretation of Components of Time Series – (Trend, Seasonal and
Cyclical Variations).
Technical Errors
EFFECT OF WRONG INTERPRETATION OF DATA
– DISTRUST OF STATISTICS
17.3.
17·3·1.
17·3·2.
17·3·3.
17·3·4.
17·3·5.
17·3·6.
17·3·6.
17·3·7.
17·4.
EXERCISE 17·1
16·1
16·1
16·2
16·2
16·2
16·3
16·3
16·5
16·7
16·11
16·12
16·15
16·19
16·20
16·21
16·22
16·24
16·26
16·27
17·1 – 17·14
17·1
17·1
17·2
17·2
17·2
17·3
17·3
17·4
17·10
17·10
17·11
17·11
17·11
(xxi)
18.
STATISTICAL DECISION THEORY
18·1.
18·2.
18·2·1.
18·2·2.
18·2·3.
18·2·4.
18·2·5.
18·2·6.
18·2·7.
18·3.
18·3·1.
18·3·2.
18·3·3.
18·3·4.
18·3·5.
18·3·6.
18·3·7.
18·3·8.
18·4.
18·4·1.
INTRODUCTION
INGREDIENTS OF DECISION PROBLEM
Acts.
States of Nature or Events.
Payoff Table.
Opportunity Loss (O.L.).
Decision Making Environment
Decision Making Under Certainty.
Decision Making Under Uncertainty.
OPTIMAL DECISION
Maximax Criterion.
Maximin Criterion.
Minimax Criterion.
Laplace Criterion of Equal Likelihoods.
Hurwicz Criterion of Realism.
Expected Monetary Value (EMV).
Expected Opportunity Loss (EOL) Criterion.
Expected Value of Perfect Information (EVPI).
DECISION TREE
Roll Back Technique of Analysing a Decision Tree.
EXERCISE 18·1.
19.
THEORY OF ATTRIBUTES
19·1.
19·2.
19·3.
19·3·1.
19·3·2.
19·3·3.
INTRODUCTION
NOTATIONS
CLASSES AND CLASS FREQUENCIES
Order of Classes and Class Frequencies.
Ultimate Class Frequency.
Relation Between Class Frequencies.
EXERCISE 19·1
19·4.
19·4·1.
19·4·2.
INCONSISTENCY OF DATA
Conditions for Consistency of Data.
Incomplete Data.
EXERCISE 19·2
19·5.
19·5·1.
19·6.
19·6·1.
19·6·2.
INDEPENDENCE OF ATTRIBUTES
Criteria of Independence of Two Attributes.
ASSOCIATION OF ATTRIBUTES
(Criterion 1). Proportion Method.
(Criterion 2). Comparison of Observed and Expected Frequencies.
18·1 – 18·35
18·1
18·2
18·2
18·2
18·2
18·3
18·3
18·4
18·4
18·5
18·5
18·6
18·6
18·6
18·7
18·10
18·11
18·12
18·23
18·24
18·29
19·1 – 19·27
19·1
19·1
19·1
19·2
19·2
19·3
19·8
19·10
19·10
19·11
19·13
19·15
19·15
19·18
19·18
19·18
(xxii)
19·6·3.
19·6·4.
(Criterion 3) Yule’s Coefficient of Association.
(Criterion 4). Coefficient of Colligation.
EXERCISE 19·3
Appendix I : NUMERICAL TABLES
Appendix II : BIBLIOGRAPHY
INDEX
19·18
19·19
19·24
T·1 – T· 9
B·1
I·1 – I·6
1
Introduction —
Meaning & Scope
1·1. ORIGIN AND DEVELOPMENT OF STATISTICS
The subject of Statistics, as it seems, is not a new discipline but it is as old as the human society itself.
It has been used right from the existence of life on this earth, although the sphere of its utility was very
much restricted. In the old days, Statistics was regarded as the ‘Science of Statecraft’ and was the byproduct of the administrative activity of the State. The word Statistics seems to have been derived from the
Latin word ‘status’ or the Italian word ‘statista’ or the German word ‘statistik’ or the French word
‘statistique’, each of which means a political state. In the ancient times the scope of Statistics was primarily
limited to the collection of the following data by the governments for framing military and fiscal policies :
(i) Age and sex-wise population of the country ;
(ii) Property and wealth of the country ;
the former enabling the government to have an idea of the manpower of the country (in order to safeguard
itself against any outside aggression) and the latter providing it with information for the introduction of new
taxes and levies.
Perhaps one of the earliest censuses of population and wealth was conducted by the Pharaohs
(Emperors) of Egypt in connection with the construction of famous ‘Pyramids’. Such censuses were later
held in England, Germany and other western countries in the middle ages. In India, an efficient system of
collecting official and administrative statistics existed even 2000 years ago - in particular during the reign
of Chandragupta Maurya (324 – 300 B.C.). Historical evidences about the prevalence of a very good system
of collecting vital statistics and registration of births and deaths even before 300 B.C. are available in
Kautilya’s ‘Arthashastra’. The records of land, agriculture and wealth statistics were maintained by
Todermal, the land and revenue minister in the reign of Akbar (1556 – 1605 A.D.). A detailed account of
the administrative and statistical surveys conducted during Akbar’s reign is available in the book ‘Ain-eAkbari’ written by Abul Fazl (in 1596 – 97), one of the nine gems of Akbar.
In Germany, the systematic collection of official statistics originated towards the end of the 18th
century when, in order to have an idea of the relative strength of different German States, information
regarding population and output—industrial and agricultural—was collected. In England, statistics were the
outcome of Napoleonic wars. The wars necessitated the systematic collection of numerical data to enable
the government to assess the revenues and expenditure with greater precision and then to levy new taxes in
order to meet the cost of war.
Sixteenth century saw the applications of Statistics for the collection of the data relating to the
movements of heavenly bodies – stars and planets – to know about their position and for the prediction of
eclipses. J. Kepler made a detailed study of the information collected by Tycko Brave (1554 – 1601)
regarding the movements of the planets and formulated his famous three laws relating to the movements of
heavenly bodies. These laws paved the way for the discovery of Newton’s law of gravitation.
Seventeenth century witnessed the origin of Vital Statistics. Captain John Graunt of London
(1620 – 1674), known as the Father of Vital Statistics, was the first man to make a systematic study of the
birth and death statistics. Important contributions in this field were also made by prominent persons like
Casper Newman (in 1691), Sir William Petty (1623 – 1687), James Dodson, Thomas Simpson and Dr.
Price. The computation of mortality tables and the calculation of expectation of life at different ages by
1·2
BUSINESS STATISTICS
these persons led to the idea of ‘Life Insurance’ and Life Insurance Institution was founded in London in
1698. William Petty wrote the book ‘Essay on Political Arithmetic’. In those days Statistics was regarded
as Political Arithmetic. This concept of Statistics as Political Arithmetic continued even in early 18th
century when J.P. Sussmilch (1707 – 1767), a Prussian Clergyman, formulated his doctrine that the ratio of
births and deaths more or less remains constant and gave statistical explanation to the theory of ‘Natural
Order of Physiocratic School’.
The backbone of the so-called modern theory of Statistics is the ‘Theory of Probability’ or the ‘Theory
of Games and Chance’ which was developed in the mid-seventeenth century. Theory of probability is the
outcome of the prevalence of gambling among the nobles of England and France while estimating the
chances of winning or losing in the gamble, the chief contributors being the mathematicians and gamblers
of France, Germany and England. Two French mathematicians Pascal (1623 – 1662) and P. Fermat
(1601 – 1665), after a lengthy correspondence between themselves ultimately succeeded in solving the
famous ‘Problem of Points’ posed by the French gambler Chevalier de-Mere and this correspondence laid
the foundation stone of the science of probability. Next stalwart in this field was, J. Bernoulli (1654 – 1705)
whose great treatise on probability ‘Ars Conjectandi’ was published posthumously in 1713, eight years
after his death by his nephew Daniel Bernoulli (1700 – 1782). This contained the famous ‘Law of Large
Numbers’ which was later discussed by Poisson, Khinchine and Kolmogorov. De-Moivre (1667 – 1754)
also contributed a lot in this field and published his famous ‘Doctrine of Chance’ in 1718 and also
discovered the Normal probability curve which is one of the most important contributions in Statistics.
Other important contributors in this field are Pierra Simon de Laplace (1749 – 1827) who published his
monumental work ‘Theoric Analytique de’s of Probabilities’, on probability in 1782; Gauss (1777 – 1855)
who gave the principle of Least Squares and established the ‘Normal Law of Errors’ independently of DeMoivre; L.A.J. Quetlet (1798 – 1874) discovered the principle of ‘Constancy of Great Numbers’ which
forms the basis of sampling; Euler, Lagrange, Bayes, etc. Russian mathematicians also have made very
outstanding contributions to the modern theory of probability, the main contributors to mention only a few
of them are : Chebychev (1821 – 1894), who founded the Russian School of Statisticians ; A. Markov
(Markov Chains) ; Liapounoff (Central Limit Theorem); A. Khinchine (Law of Large Numbers) ; A
Kolmogorov (who axiomised the calculus of probability) ; Smirnov, Gnedenko and so on.
Modern stalwarts in the development of the subject of Statistics are Englishmen who did pioneering
work in the application of Statistics to different disciplines. Francis Galton (1822 – 1921) pioneered the
study of ‘Regression Analysis’ in Biometry; Karl Pearson (1857 – 1936) who founded the greatest statistical
laboratory in England pioneered the study of ‘Correlation Analysis’. His Chi-Square test (χ2-test) of
Goodness of Fit is the first and most important of the tests of significance in Statistics ; W.S. Gosset with
his t-test ushered in an era of exact (small) sample tests. Perhaps most of the work in the statistical theory
during the past few decades can be attributed to a single person Sir Ronald A. Fisher (1890 – 1962) who
applied Statistics to a variety of diversified fields such as genetics, biometry, psychology and education,
agriculture, etc., and who is rightly termed as the Father of Statistics. In addition to enhancing the existing
statistical theory he is the pioneer in Estimation Theory (Point Estimation and Fiducial Inference); Exact
(small) Sampling Distributions ; Analysis of Variance and Design of Experiments. His contributions to the
subject of Statistics are described by one writer in the following words :
“R.A. Fisher is the real giant in the development of the theory of Statistics.”
It is only the varied and outstanding contributions of R.A. Fisher that put the subject of Statistics on a
very firm footing and earned for it the status of a full-fledged science.
Indian statisticians also did not lag behind in making significant contributions to the development of
Statistics in various diversified fields. The valuable contributions of C.R. Rao (Statistical Inference);
Parthasarathy (Theory of Probability); P.C. Mahalanobis and P.V. Sukhatme (Sample Surveys) ; S.N. Roy
(Multivariate Analysis) ; R.C. Bose, K.R. Nair, J.N. Srivastava (Design of Experiments), to mention only a
few, have placed India’s name in the world map of Statistics.
1·2. DEFINITION OF STATISTICS
Statistics has been defined differently by different writers from time to time so much so that scholarly
articles have collected together hundreds of definitions, emphasizing precisely the meaning, scope and
limitations of the subject. The reasons for such a variety of definitions may be broadly classified as follows :
INTRODUCTION — MEANING & SCOPE
1·3
(i) The field of utility of Statistics has been increasing steadily and thus different people defined it
differently according to the developments of the subject. In old days, Statistics was regarded as the ‘science
of statecraft’ but today it embraces almost every sphere of natural and human activity. Accordingly, the old
definitions which were confined to a very limited and narrow field of enquiry were replaced by the new
definitions which are more exhaustive and elaborate in approach.
(ii) The word Statistics has been used to convey different meanings in singular and plural sense. When
used as plural, statistics means numerical set of data and when used in singular sense it means the science
of statistical methods embodying the theory and techniques used for collecting, analysing and drawing
inferences from the numerical data.
It is practically impossible to enumerate all the definitions given to Statistics both as ‘Numerical Data’
and ‘Statistical Methods’ due to limitations of space. However, we give below some selected definitions.
1.
2.
3.
4.
WHAT THEY SAY ABOUT STATISTICS — SOME DEFINITIONS
”STATISTICS AS NUMERICAL DATA”
“Statistics are the classified facts representing the conditions of the people in a
State…specially those facts which can be stated in number or in tables of numbers or in
any tabular or classified arrangement.”—Webster.
“Statistics are numerical statements of facts in any department of enquiry placed in
relation to each other.”— Bowley.
“By statistics we mean quantitative data affected to a marked extent by multiplicity of
causes”.—Yule and Kendall.
“Statistics may be defined as the aggregate of facts affected to a marked extent by
multiplicity of causes, numerically expressed, enumerated or estimated according to a
reasonable standard of accuracy, collected in a systematic manner, for a predetermined
purpose and placed in relation to each other.”—Prof. Horace Secrist.
Remarks and Comments. 1. According to Webster’s definition only numerical facts can be termed
Statistics. Moreover, it restricts the domain of Statistics to the affairs of a State i.e., to social sciences. This
is a very old and narrow definition and is inadequate for modern times since today, Statistics embraces all
sciences – social, physical and natural.
2. Bowley’s definition is more general than Webster’s since it is related to numerical data in any
department of enquiry. Moreover it also provides for comparative study of the figures as against mere
classification and tabulation of Webster’s definition.
3. Yule and Kendall’s definition refers to numerical data affected by a multiplicity of causes. This is
usually the case in social, economic and business phenomenon. For example, the prices of a particular
commodity are affected by a number of factors viz., supply, demand, imports, exports, money in
circulation, competitive products in the market and so on. Similarly, the yield of a particular crop depends
upon multiplicity of factors like quality of seed, fertility of soil, method of cultivation, irrigation facilities,
weather conditions, fertilizer used and so on.
4. Secrist’s definition seems to be the most exhaustive of all the four. Let us try to examine it in details.
(i) Aggregate of Facts. Simple or isolated items cannot be termed as Statistics unless they are a part of
aggregate of facts relating to any particular field of enquiry. For instance, the height of an individual or the
price of a particular commodity do not form Statistics as such figures are unrelated and uncomparable.
However, aggregate of the figures of births, deaths, sales, purchase, production, profits, etc., over different
times, places, etc., will constitute Statistics.
(ii) Affected by Multiplicity of Causes. Numerical figures should be affected by multiplicity of
factors. This point has already been elaborated in remark 3 above. In physical sciences, it is possible to
isolate the effect of various factors on a single item but it is very difficult to do so in social sciences,
particularly when the effect of some of the factors cannot be measured quantitatively. However, statistical
techniques have been devised to study the joint effect of a number of a factors on a single item (Multiple
Correlation) or the isolated effect of a single factor on the given item (Partial Correlation) provided the
effect of each of the factors can be measured quantitatively.
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(iii) Numerically Expressed. Only numerical data constitute Statistics. Thus the statements like ‘the
standard of living of the people in Delhi has improved’ or ‘the production of a particular commodity is
increasing’ do not constitute Statistics. In particular, the qualitative characteristics which cannot be
measured quantitatively such as intelligence, beauty, honesty, etc., cannot be termed as Statistics unless
they are numerically expressed by assigning particular scores as quantitative standards. For example,
intelligence is not Statistics but the intelligence quotients which may be interpreted as the quantitative
measure of the intelligence of individuals could be regarded as Statistics.
(iv) Enumerated or Estimated According to Reasonable Standard of Accuracy. The numerical
data pertaining to any field of enquiry can be obtained by completely enumerating the underlying
population. In such a case data will be exact and accurate (but for the errors of measurement, personal bias,
etc.). However, if complete enumeration of the underlying population is not possible (e.g., if population is
infinite, or if testing is destructive i.e., if the item is destroyed in the course of inspection just like in testing
explosives, light bulbs, etc.), and even if possible it may not be practicable due to certain reasons (such as
population being very large, high cost of enumeration per unit and our resources being limited in terms of
time and money, etc.), then the data are estimated by using the powerful techniques of Sampling and
Estimation theory. However, the estimated values will not be as precise and accurate as the actual values.
The degree of accuracy of the estimated values largely depends on the nature and purpose of the enquiry.
For example, while measuring the heights of individuals accuracy will be aimed in terms of fractions of an
inch whereas while measuring distance between two places it may be in terms of metres and if the places
are very distant, e.g., say Delhi and London, the difference of few kilometres may be ignored. However,
certain standards of accuracy must be maintained for drawing meaningful conclusions.
(v) Collected in a Systematic Manner. The data must be collected in a very systematic manner. Thus,
for any socio-economic survey, a proper schedule depending on the object of enquiry should be prepared
and trained personnel (investigators) should be used to collect the data by interviewing the persons. An
attempt should be made to reduce the personal bias to the minimum. Obviously, the data collected in a
haphazard way will not conform to the reasonable standards of accuracy and the conclusions based on them
might lead to wrong or misleading decisions.
(vi) Collected for a Pre-determined Purpose. It is of utmost importance to define in clear and
concrete terms the objectives or the purpose of the enquiry and the data should be collected keeping in view
these objectives. An attempt should not be made to collect too many data some of which are never
examined or analysed i.e., we should not waste time in collecting the information which is irrelevant for our
enquiry. Also it should be ensured that no essential data are omitted. For example, if the purpose of enquiry
is to measure the cost of living index for low income group people, we should select only those
commodities or items which are consumed or utilised by persons belonging to this group. Thus for such an
index, the collection of the data on the commodities like scooters, cars, refrigerators, television sets, high
quality cosmetics, etc., will be absolutely useless.
(vii) Comparable. From practical point of view, for statistical analysis the data should be comparable.
They may be compared with respect to some unit, generally time (period) or place. For example, the data
relating to the population of a country for different years or the population of different countries in some
fixed year constitute Statistics, since they are comparable. However, the data relating to the size of the shoe
of an individual and his intelligence quotient (I.Q.) do not constitute Statistics as they are not comparable.
In order to make valid comparisons the data should be homogeneous i.e., they should relate to the same
phenomenon or subject.
5. From the definition of Horace Secrist and its discussion in remark 4 above, we may conclude that :
“All Statistics are numerical statements of facts but all numerical statements of facts are not
Statistics”.
6. We give below the definitions of Statistics used in singular sense i.e., Statistics as Statistical
Methods.
WHAT THEY SAY ABOUT STATISTICS — SOME DEFINITIONS
“STATISTICS AS STATISTICAL METHODS”
1. Statistics may be called the science of counting. —Bowley A.L .
2. Statistics may rightly be called the science of averages. —Bowley A.L .
INTRODUCTION — MEANING & SCOPE
1·5
3. Statistics is the science of the measurement of social organism, regarded as a whole in all
its manifestations. —Bowley A.L .
4. “Statistics is the science of estimates and probabilities.” — Boddington
5. “The science of Statistics is the method of judging collective, natural or social phenomenon
from the results obtained from the analysis or enumeration or collection of estimates.”
—King
6. Statistics is the science which deals with classification and tabulation of numerical facts as
the basis for explanation, description and comparison of phenomenon.”—Lovin
7. “Statistics is the science which deals with the methods of collecting, classifying, presenting,
comparing and interpreting numerical data collected to throw some light on any sphere of
enquiry.”—Selligman
8. “Statistics may be defined as the science of collection, presentation, analysis and
interpretation of numerical data.” —Croxton and Cowden
9. “Statistics may be regarded as a body of methods for making wise decisions in the face of
uncertainty.”—Wallis and Roberts
10. “Statistics is a method of decision making in the face of uncertainty on the basis of
numerical data and calculated risks.”—Prof. Ya-Lun-Chou
11. “The science and art of handling aggregate of facts—observing, enumeration, recording,
classifying and otherwise systematically treating them.”—Harlow
Some Comments and Remarks. 1. The first three definitions due to Bowley are inadequate.
2. Boddington’s definition also fails to describe the meaning and functions of Statistics since it is
confined to only probabilities and estimates which form only a part of the modern statistical tools and do
not describe the science of Statistics in all its manifestations.
3. King’s definition is also inadequate since it confines Statistics only to social sciences. Lovitt’s
definition is fairly satisfactory, though incomplete. Selligman’s definition, though very short and simple is
quite comprehensive. However, the best of all the above definitions seems to be one given by Croxton and
Cowden.
4. Wallis and Roberts’ definition is quite modern since statistical methods enable us to arrive at valid
decisions. Prof. Chou’s definition in number 10 is a modified form of this definition.
5. Harlow’s definition describes Statistics both as a science and an art—science, since it provides tools
and laws for the analysis of the numerical information collected from the source of enquiry and art, since it
undeniably has its basis upon numerical data collected with a view to maintain a particular balance and
consistency leading to perfect or nearly perfect conclusions. A statistician like an artist will fail in his job if
he does not possess the requisite skill, experience and patience while using statistical tools for any problem.
1·3. IMPORTANCE AND SCOPE OF STATISTICS
In the ancient times Statistics was regarded only as the science of Statecraft and was used to collect
information relating to crimes, military strength, population, wealth, etc., for devising military and fiscal
policies. But with the concept of Welfare State taking roots almost all over the world, the scope of Statistics
has widened to social and economic phenomenon. Moreover, with the developments in the statistical
techniques during the last few decades, today, Statistics is viewed not only as a mere device for collecting
numerical data but as a means of sound techniques for their handling and analysis and drawing valid
inferences from them. Accordingly, it is not merely a by-product of the administrative set up of the State
but it embraces all sciences—social, physical, and natural, and is finding numerous applications in various
diversified fields such as agriculture, industry, sociology, biometry, planning, economics, business,
management, psychometry, insurance, accountancy and auditing, and so on. It is rather impossible to think
of any sphere of human activity where Statistics does not creep in. It will not be exaggeration to say that
Statistics has assumed unprecedented dimensions these days and statistical thinking is becoming more and
more indispensable every day for an able citizenship. In fact to a very striking degree, the modern culture
has become a statistical culture and the subject of Statistics has acquired tremendous progress in the recent
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past so much so that an elementary knowledge of statistical methods has become a part of the general
education in the curricula of many universities all over the world. The importance of Statistics is amply
explained in the following words of Carrol D. Wright (1887), United States Commissioner of the Bureau of
Labour :
“To a very striking degree our culture has become a Statistical culture. Even a person who may never
have heard of an index number is affected…by … of those index numbers which describe the cost of living.
It is impossible to understand Psychology, Sociology, Economics, Finance or a Physical Science without
some general idea of the meaning of an average, of variation, of concomitance, of sampling, of how to
interpret charts and tables.”
There is no ground for misgivings regarding the practical realisation of the dream of H.G. Wells viz.,
“Statistical thinking will one day be as necessary for effective citizenship as the ability to read and write.”
Statistics has become so much indispensable in all phases of human endeavour that it is often remarked,
“Statistics is what statisticians do” and it appears that Bowley was right when he said, “A knowledge of
Statistics is like a knowledge of foreign language or of algebra; it may prove of use at any time under any
circumstances.”
Let us now discuss briefly the importance of Statistics in some different disciplines.
Statistics in Planning. Statistics is imdispensable in planning – may it be in business, economics or
government level. The modern age is termed as the ‘age of planning’ and almost all organisations in the
government or business or management are resorting to planning for efficient working and for formulating
policy decisions. To achieve this end, the statistical data relating to production, consumption, prices,
investment, income, expenditure and so on and the advanced statistical techniques such as index numbers,
time series analysis, demand analysis and forecasting techniques for handling such data are of paramount
importance. Today efficient planning is a must for almost all countries, particularly the developing
economies for their economic development and in order that planning is successful, it must be based on a
correct and sound analysis of complex statistical data. For instance, in formulating a five-year plan, the
government must have an idea of the age and sex-wise break up of the population projections of the country
for the next five years in order to develop its various sectors like agriculture, industry, textiles, education
and so on. This is achieved through the powerful statistical tool of forecasting by making use of the
population data for the previous years. Even for making decisions concerning the day to day policy of the
country, an accurate statistical knowledge of the age and sex-wise composition of the population is
imperative for the government. In India, the use of Statistics in planning was well visualised long back and
the National Sample Survey (N.S.S.) was primarily set up in 1950 for the collection of statistical data for
planning in India.
Statistics in State. As has already been pointed out, in the old days Statistics was the science of Statecraft and its objective was to collect data relating to manpower, crimes, income and wealth, etc., for
formulating suitable military and fiscal policies. With the inception of the idea of Welfare State and its
taking deep roots in almost all the countries, today statistical data relating to prices, production,
consumption, income and expenditure, investments and profits, etc., and statistical tools of index numbers,
time series analysis, demand analysis, forecasting, etc., are extensively used by the governments in
formulating economic policies. (For details see Statistics in Economics). Moreover as pointed out earlier
(Statistics in planning), statistical data and techniques are indispensable to the government for planning
future economic programmes. The study of population movement i.e., population estimates, population
projections and other allied studies together with birth and death statistics according to age and sex
distribution provide any administration with fundamental tools which are indispensable for overall planning
and evaluation of economic and social development programmes. The facts and figures relating to births,
deaths and marriages are of extreme importance to various official agencies for a variety of administrative
purposes. Mortality (death) statistics serve as a guide to the health authorities for sanitary improvements,
improved medical facilities and public cleanliness. The data on the incidence of diseases together with the
number of deaths by age and nature of diseases are of paramount importance to health authorities in taking
appropriate remedial action to prevent or control the spread of the disease. The use of statistical data and
statistical techniques is so wide in government functioning that today, almost all ministries and the
departments in the government have a separate statistical unit. In fact, today, in most countries the State
(government) is the single unit which is the biggest collector and user of statistical data. In addition to the
INTRODUCTION — MEANING & SCOPE
1·7
various statistical bureaux in all the ministries and the government departments in the Centre and the States,
the main Statistical Agencies in India are Central Statistical Organisation (C.S.O.) ; National Sample
Survey (N.S.S.), now called National Sample Survey Organisation (N.S.S.O.) and the Registrar General of
India (R.G.I.).
Statistics in Economics. In old days, Economic Theories were based on deductive logic only.
Moreover, the statistical techniques were not that much advanced for applications in other disciplines. It
gradually dawned upon economists of the Deductive School to use Statistics effectively by making
empirical studies.
In 1871, W.S. Jevons, wrote that :
“The deductive science of economy must be verified and rendered useful from the purely inductive
science of Statistics. Theory must be invested with the reality of life and fact.”
These views were supported by Roscher, Kines and Hildebrand of the Historical School (1843 – 1883),
Alfred Marshall, Pareto, Lord Keynes. The following quotation due to Prof. Alfred Marshall in 1890 amply
illustrates the role of Statistics in Economics :
“Statistics are the straws out of which I, like every other economist, have to make bricks.”
Statistics plays a very vital role in Economics so much so that in 1926, Prof. R.A. Fisher complained of
“the painful misapprehension that Statistics is a branch of Economics.”
Statistical data and advanced techniques of statistical analysis have proved immensely useful in the
solution of a variety of economic problems such as production, consumption, distribution of income and
wealth, wages, prices, profits, savings, expenditure, investment, unemployment, poverty, etc. For example,
the studies of consumption statistics reveal the pattern of the consumption of the various commodities by
different sections of the society and also enable us to have some idea about their purchasing capacity and
their standard of living. The studies of production statistics enable us to strike a balance between supply
and demand which is provided by the laws of supply and demand. The income and wealth statistics are
mainly helpful in reducing the disparities of income. The statistics of prices are needed to study the price
theories and the general problem of inflation through the construction of the cost of living and wholesale
price index numbers. The statistics of market prices, costs and profits of different individual concerns are
needed for the studies of competition and monopoly. Statistics pertaining to some macro-variables like
production, income, expenditure, savings, investments, etc., are used for the compilation of National
Income Accounts which are indispensable for economic planning of a country. Exchange statistics reflect
upon the commercial development of a nation and tell us about the money in circulation and the volume of
business done in the country. Statistical techniques have also been used in determining the measures of
Gross National Product and Input-Output Analysis. The advanced and sound statistical techniques have
been used successfully in the analysis of cost functions, production functions and consumption functions.
Use of Statistics in Economics has led to the formulation of many economic laws some of which are
mentioned below for illustration :
A detailed and systematic study of the family budget data which gives a detailed account of the family
budgets showing expenditure on the main items of family consumption together with family structure and
composition, family income and various other social, economic and demographic characteristics led to the
famous Engel’s Law of Consumption in 1895. Vilfredo Pareto in 19th-20th century propounded his famous
Law of Distribution of Income by making an empirical study of the income data of various countries of the
world at different times. The study of the data pertaining to the actual observation of the behaviour of
buyers in the market resulted in the Revealed Preference Analysis of Prof. Samuelson.
Time Series Analysis, Index Numbers, Forecasting Techniques and Demand Analysis are some of the
very powerful statistical tools which are used immensely in the analysis of economic data and also for
economic planning. For instance, time series analysis is extremely used in Business and Economic
Statistics for the study of the series relating to prices, production and consumption of commodities, money
in circulation, bank deposits and bank clearings, sales in a departmental store, etc.,
(i) to identify the forces or components at work, the net effect of whose interaction is exhibited by the
movement of the time series;
(ii) to isolate, study, analyse and measure them independently.
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BUSINESS STATISTICS
The index numbers which are also termed as ‘economic barometers’ are the numbers which reflect the
changes over specified period of time in (i) prices of different commodities, (ii) industrial/agricultural
production, (iii) sales, (iv) imports and exports, (v) cost of living, etc., and are extremely useful in economic
planning. For instance, the cost of living index numbers are used for (i) the calculation of real wages and
for determining the purchasing power of the money; (ii) the deflation of income and value series in national
accounts; (iii) grant of dearness allowance (D.A.) or bonus to the workers in order to enable them to meet
the increased cost of living and so on.
The demand analysis consists in making an economic study of the market data to determine the relation
between :
(i) the prices of a given commodity and its absorption capacity for the market i.e., demand; and
(ii) the price of a commodity and its output i.e., supply.
Forecasting techniques based on the method of curve fitting by the principle of least squares and
exponential smoothing are indispensable tools for economic planning.
The increasing interaction of mathematics and statistics with economics led to the development of a
new discipline called Econometrics—and the first Econometric Society was founded in U.S.A. in 1930 for
“the advancement of economic theory in its relation to mathematics and statistics…” Econometrics aimed
at making Economics a more realistic, precise, logical and practical science. Econometric models based on
sound statistical analysis are used for maximum exploitation of the available resources. In other words, an
attempt is made to obtain optimum results subject to a number of constraints on the resources at our
disposal, say, of production capacity, capital, technology, precision, etc., which are determined statistically.
Statistics in Business and Management. Prior to the Industrial Revolution, when the production was
at the handicraft stage, the business activities were very much limited and were confined only to small units
operating in their own areas. The owner of the concern personally looked after all the departments of
business activity like sales, purchase, production, marketing, finance and so on. But after the Industrial
Revolution, the developments in business activities have taken such unprecedented dimensions both in the
size and the competition in the market that the activities of most of the business enterprises and firms are
confined not only to one particular locality, town or place but to larger areas. Some of the leading houses
have the network of their business activities in almost all the leading towns and cities of the country and
even abroad. Accordingly it is impossible for a single person (the owner of the concern) to look after its
activities and management has become a specialised job. The manager and a team of management
executives is imperative for the efficient handling of the various operations like sales, purchase, production,
marketing, control, finance, etc., of the business house. It is here that statistical data and the powerful
statistical tools of probability, expectation, sampling techniques, tests of significance, estimation theory,
forecasting techniques and so on play an indispensable role. According to Wallis and Roberts : “Statistics
may be regarded as a body of methods for making wise decisions in the face of uncertainty.” A refinement
over this definition is provided by Prof. Ya-Lun-Chou as follows : “Statistics is a method of decision
making in the face of uncertainty on the basis of numerical data and calculated risks.” These definitions
reflect the applications of Statistics in business since modern business has its roots in the accuracy and
precision of the estimates and statistical forecasting regarding the future demand for the product, market
trends and so on. Business forecasting techniques which are based on the compilation of useful statistical
information on lead and lag indicators are very useful for obtaining estimates which serve as a guide to
future economic events. Wrong expectations which might be the result of faulty and inaccurate analysis of
various factors affecting a particular phenomenon might lead to his disaster. The time series analysis is a
very important statistical tool which is used in business for the study of :
(i) Trend (by method of curve fitting by the principle of least squares) in order to obtain the estimates
of the probable demand of the goods; and
(ii) Seasonal and Cyclical movements in the phenomenon, for determining the ‘Business Cycle’ which
may also be termed as the four-phase cycle composed of prosperity (period of boom), recession, depression
and recovery. The upswings and downswings in business depend on the cumulative nature of the economic
forces (affecting the equilibrium of supply and demand) and the interaction between them. Most of the
business and commercial series e.g., series relating to prices, production, consumption, profits, investments,
wages, etc., are affected to a great extent by business cycles. Thus the study of business cycles is of
INTRODUCTION — MEANING & SCOPE
1·9
paramount importance in business and a businessman who ignores the effects of booms and depression is
bound to fail since his estimates and forecasts will definitely be faulty.
The studies of Economic Barometers (Index Numbers of Prices) enable the businessman to have an
idea about the purchasing power of money. The statistical tools of demand analysis enable the businessman
to strike a balance between supply and demand. [For details, see Statistics in Economics].
The technique of Statistical Quality Control, through the powerful tools of ‘Control Charts’ and
‘Inspection Plans’ is indispensable to any business organisation for ensuring that the quality of the
manufactured product is in conformity with the consumer’s specifications. (For details see Statistics in
Industry).
Statistical tools are used widely by business enterprises for the promotion of new business. Before
embarking upon any production process, the business house must have an idea about the quantum of the
product to be manufactured, the amount of the raw material and labour needed for it, the quality of the
finished product, marketing avenues for the product, the competitive products in the market and so on. Thus
the formulation of a production plan is a must and this cannot be achieved without collecting the statistical
information on the above items without resorting to the powerful technique of ‘Sample Surveys’. As such,
most of the leading business and industrial concerns have full-fledged statistical units with trained and
efficient statisticians for formulating such plans and arriving at valid decisions in the face of uncertainty
with calculated risks. These units also carry on research and development programmes for the improvement
of the quality of the existing products (in the light of the competitive products in the market), introduction
of new products and optimisation of the profits with existing resources at their disposal.
Statistical tools of probability and expectation are extremely useful in Life Insurance which is one of
the pioneer branches of Business and Commerce to use Statistics since the end of the seventeenth century.
Statistical techniques have also been used very widely by business organisations in :
(i) Marketing Decisions (based on the statistical analysis of consumer preference studies – demand
analysis).
(ii) Investment (based on sound study of individual shares and debentures).
(iii) Personnel Administration (for the study of statistical data relating to wages, cost of living,
incentive plans, effect of labour dispute/unrest on the production, performance standards, etc.).
(iv) Credit policy.
(v) Inventory Control (for co-ordination between production and sales).
(vi) Accounting (for evaluation of the assets of the business concerns). (For details see Statistics in
Accountancy and Auditing).
(vii) Sales Control (through the statistical data pertaining to market studies, consumer preference
studies, trade channel studies and readership surveys, etc.), and so on.
From the above discussion it is obvious that the use of statistical data and techniques is indispensable
in almost all the branches of business activity.
Statistics in Accountancy and Auditing. Today, the science of Statistics has assumed such
unprecedented dimensions that even the subjects like Accountancy and Auditing have not escaped its
domain. The ever-increasing applications of the statistical data and the advanced statistical techniques in
Accountancy and Auditing are well supported by the inclusion of a compulsory paper on Statistics both in
the Chartered Accountants (Foundation) and Cost and Works Accountants (Intermediate) examinations
curriculum. Statistics has innumerable applications in accountancy and auditing. For example the statistical
data on some macro-variables like income, expenditure, investment, profits, production, savings, etc., are
used for the compilation of National Income Accounts which provide information on the value added by
different sectors of economy and are very helpful in formulating economic policies. The statistical study
(Correlation Analysis) of profit and dividend statistics enables one to predict the probable dividends for the
future years. Further, in Accountancy, the statistics of assets and liabilities, and income and expenditure are
helpful to ascertain the financial results of various operations.
A very important application of Statistics in accountancy is in the ‘Method of Inflation Accounting’
which consists in revaluating the accounting records based on historical costs of assets after adjusting for
the changes in the purchasing power of money. This is achieved through the powerful statistical tools of
Price Index Numbers and Price Deflators.
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The Regression Analysis theory is of immense help in Cost Accounting in forecasting cost or price for
any given value of the dependent variable. Suppose there exists a functional relation between cost of
production (c) and the price of the product (p), of the form :
c = f (p)
Then with the statistical tools of regression analysis we can predict the effect of changes in future
prices on the cost of production. Statistical techniques are also greatly used in forecasting profits,
determination of trend, computation of financial and other ratios, cost-volume-profit analysis and so on.
The efficacy of the implementation of a new investment plan can be tested by using the statistical tests of
significance.
In Auditing, sampling techniques are used widely for test checking. The business transactions and the
volumes of the various items comprising balances in various accounts are so heavy (voluminous) that it is
practically impossible to resort to 100% examination and analysis of the records because of limitations of
time, money and staff at our disposal. Accordingly, sampling techniques based on sound statistical and
scientific reasoning are used effectively to examine thoroughly only a sample (fraction – 2% or 5%) of the
transactions or the items comprising a balance and drawing inferences about the whole lot (data) by using
statistical techniques of Estimation and Inference.
Statistics in Industry. In industry, Statistics is extensively used in ‘Quality Control’. The main
objective in any production process it to control the quality of the manufactured product so that it conforms
to specifications. This is called process control and is achieved through the powerful technique of
control charts and inspection plans. The discovery of the control charts was made by a young physicist
Dr. W.A. Shewhart of the Bell Telephone Laboratories (U.S.A.) in 1924 and the following years and is
based on setting the 3σ (3 – sigma) control limits which has its basis on the theory of probability and
normal distribution. Inspection plans are based on special kind of sampling techniques which are a very
important aspect of statistical theory.
Statistics in Physical Sciences. The applications of Statistics in Astronomy, which is a physical
science, have already been discussed above. In physical sciences, a large number of measurements are
taken on the same item. There is bound to be variation in these measurements. In order to have an idea
about the degree of accuracy achieved, the statistical techniques (Interval Estimation – confidence intervals
and confidence limits) are used to assign certain limits within which the true value of the phenomenon may
be expected to lie. The desire for precision was first felt in physical sciences and this led the science to
express the facts under study in quantitative form. The statistical theory with the powerful tools of
sampling, estimation (point and interval), design of experiments, etc., is most effective for the analysis of
the quantitative expression of all fields of study. Today, there is an increasing use of Statistics in most of
the physical sciences such as astronomy, geology, engineering, physics and meteorology.
Statistics in Social Sciences. According to Bowley, “Statistics is the science of the measurement of
social organism, regarded as a whole in all its manifestations.” In the words of W.I. King, “The science of
Statistics is the method of judging collective, natural or social phenomenon from the results obtained from
the analysis or enumeration or collection of estimates.” These words of Bowley and King amply reflect
upon the importance of Statistics in social sciences.
Every social phenomenon is affected to a marked extent by a multiplicity of factors which bring out the
variation in observations from time to time, place to place and object to object. Statistical tools of
Regression and Correlation Analysis can be used to study and isolate the effect of each of these factors on
the given observation. Sampling Techniques and Estimation Theory are very powerful and indispensable
tools for conducting any social survey, pertaining to any strata of society and then analysing the results and
drawing valid inferences. The most important application of statistics in sociology is in the field of
Demography for studying mortality (death rates), fertility (birth rates), marriages, population growth and so
on. In fact statistical data and statistical techniques have been used so frequently and in so many problems
in social sciences that Croxton and Cowden have remarked :
“Without an adequate understanding of the statistical methods, the investigator in the social sciences
may be like the blind man groping in a dark room for a black cat that is not there. The methods of Statistics
are useful in an over-widening range of human activities in any field of thought in which numerical data
may be had.”
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1·11
Statistics in Biology and Medical Sciences. Sir Francis Galton (1822 – 1911), a British Biometrician
pioneered the use of statistical methods with his work on ‘Regression’ in connection with the inheritance of
stature. According to Prof. Karl Pearson (1857 – 1936) who pioneered the study of ‘Correlation Analysis’,
the whole theory of heredity rests on statistical basis. In his Grammar of Sciences he says, “The whole
problem of evolution is a problem of vital statistics, a problem of longevity, of fertility, of health, of disease
and it is impossible for the evolutionist to proceed without statistics as it would be for the Registrar
General to discuss the rational mortality without an enumeration of the population, a classification of
deaths and a knowledge of statistical theory.”
In medical sciences also, the statistical tools for the collection, presentation and analysis of observed
factual data relating to the causes and incidence of diseases are of paramount importance. For example, the
factual data relating to pulse rate, body temperature, blood pressure, heart beats, weight, etc., of the patient
greatly help the doctor for the proper diagnosis of the disease; statistical papers are used to study heart beats
through electro-cardiogram (E.C.G.). Perhaps the most important application of Statistics in medical
sciences lies in using the tests of significance (more precisely Student’s t-test) for testing the efficacy of a
manufacturing drug, injection or medicine for controlling/curing specific ailments. The testing of the
effectiveness of a medicine by the manufacturing concern is a must, since only after the effectiveness of the
medicine is established by the sound statistical techniques that it will venture to manufacture it on a large
scale and bring it out in the market. Comparative studies for the effectiveness of different medicines by
different concerns can also be made by statistical techniques.
Statistics in Psychology and Education. Statistics has been used very widely in education and
psychology too e.g., in the scaling of mental tests and other psychological data; for measuring the reliability
and validity of test scrores ; for determing the Intelligence Quotient (I.Q.) ; in Item Analysis and Factor
Analysis. The vast applications of statistical data and statistical theories have given rise to a new discipline
called ‘Psychometry’.
1·4. LIMITATIONS OF STATISTICS
Although Statistics is indispensable to almost all sciences—social, physical and natural, and is very
widely used in almost all spheres of human activity, it is not without limitations which restrict its scope and
utility.
1. Statistics does not study qualitative phenomenon. ‘Statistics are numerical statements in any
department of enquiry placed in relation to each other’. Since Statistics is a science dealing with a set of
numerical data, it can be applied to the study of only those phenomena which can be measured
quantitatively. Thus the statements like ‘population of India has increased considerably during the last few
years’ or ‘the standard of living of the people in Delhi has gone up as compared with last year’, do not
constitute Statistics. As such Statistics cannot be used directly for the study of quality characteristics like
health, beauty, honesty, welfare, poverty, etc., which cannot be measured quantitatively. However, the
techniques of statistical analysis can be applied to qualitative phenomena indirectly by expressing them
numerically after assigning particular scores or quantitative standards. For instance, attribute of intelligence
in a group of individuals can be studied on the basis of their intelligence quotients (I.Q.’s) which may be
regarded as the quantitative measure of the individuals’ intelligence.
2. Statistics does not study individuals. According to Prof. Horace Secrist, “By Statistics we mean
aggregate of facts affected to a marked extent by multiplicity of factors…and placed in relation to each
other.” Thus a single or isolated figure cannot be regarded as Statistics unless it is a part of the aggregate of
facts relating to any particular field of enquiry. Thus statistical methods do not give any recognition to an
object or a person or an event in isolation. This is a serious limitation of Statistics. For instance, the price of
a single commodity, the profit of a particular concern or the production of a particular business house do
not constitute statistics since these figures are unrelated and uncomparable. However, the aggregate of
figures relating to prices and consumption of various commodities, the sales and profits of a business
house, the income, expenditure, production, etc., over different periods of time, places, etc., will be
Statistics. Thus from statistical point of view the figure of the population of a particular country in some
given year is useless unless we are also given the figures of the population of the country for different years
or of different countries for the same year for comparative studies. Hence Statistics is confined only to
those problems where group characteristics are to be studied.
1·12
BUSINESS STATISTICS
3. Statistical laws are not exact. Since the statistical laws are probabilistic in nature, inferences based
on them are only approximate and not exact like the inferences based on mathematical or scientific
(physical and natural sciences) laws. Statistical laws are true only on the average. If the probability of
getting a head in a single throw of a coin is 21 , it does not imply that if we toss a coin 10 times, we shall get
five heads and five tails. In 10 throws of a coin we may get 8 heads, 9 heads or all the 10 heads, or we may
not get even a single head. By this we mean that if the experiment of throwing the coin is carried on
indefinitely (very large number of times), then we should expect on the average 50% heads and 50% tails.
4. Statistics is liable to be misused. Perhaps the most significant limitation of Statistics is that it must
be used by experts. According to Bowley, “Statistics only furnishes a tool though imperfect which is
dangerous in the hands of those who do not know its use and deficiencies.” Statistical methods are the most
dangerous tools in the hands of the inexperts. Statistics is one of those sciences whose adepts must exercise
the self-restraint of an artist. The greatest limitation of Statistics is that it deals with figures which are
innocent in themselves and do not bear on their face the label of their quality and can be easily distorted,
manipulated or moulded by politicians, dishonest or unskilled workers, unscrupulous people for personal
selfish motives. Statistics neither proves nor disproves anything. It is merely a tool which, if rightly used
may prove extremely useful but if misused by inexperienced, unskilled and dishonest statisticians might
lead to very fallacious conclusions and even prove to be disastrous. In the words of W.I. King, “Statistics
are like clay of which you can make a God or a Devil as you please.” At another place he remarks, “Science
of Statistics is the useful servant but only of great value to those who understand its proper use.”
Thus the use of Statistics by the experts who are well experienced and skilled in the analysis and
interpretation of statistical data for drawing correct and valid inferences very much reduces the chances of
mass popularity of this important science.
1·5. DISTRUST OF STATISTICS
The improper use of statistical tools by unscrupulous people with an improper statistical bend of mind
has led to the public distrust in Statistics. By this we mean that public loses its belief, faith and confidence
in the science of Statistics and starts condemning it. Such irresponsible, inexperienced and dishonest
persons who use statistical data and statistical techniques to fulfill their selfish motives have discredited the
science of Statistics with some very interesting comments, some of which are stated below :
(i) An ounce of truth will produce tons of Statistics.
(ii) Statistics can prove anything.
(iii) Figures do not lie. Liars figure.
(iv) Statistics is an unreliable science.
(v) There are three types of lies – lies, damned lies and Statistics, wicked in the order of their
naming ; and so on.
Some of the reasons for the above remarks may be enumerated as follows :
(a) Figures are innocent and believable, and the facts based on them are psychologically more
convincing. But it is a pity that figures do not have the label of quality on their face.
(b) Arguments are put forward to establish certain results which are not true by making use of
inaccurate figures or by using incomplete data, thus distorting the truth.
(c) Though accurate, the figures might be moulded and manipulated by dishonest and unscrupulous
persons to conceal the truth and present a working and distorted picture of the facts to the public for
personal and selfish motives.
Hence, if Statistics and its tools are misused, the fault does not lie with the science of Statistics. Rather,
it is the people who misuse it, are to be blamed.
Utmost care and precautions should be taken for the interpretation of statistical data in all its
manifestations. “Statistics should not be used as a blind man uses a lamp-post for support instead of
illumination.” However, there are misapprehensions about the argument that Statistics can be used
effectively by expert statisticians, as is given in the following remark due to Wallis and Roberts:
“He who accepts statistics indiscriminately will often be duped unnecessarily. But he who distrusts
statistics indiscriminately will often be ignorant unnecessarily. There is an accessible alternative between
INTRODUCTION — MEANING & SCOPE
1·13
blind gullibility and blind distrust. It is possible to interpret statistics skillfully. The art of interpretation
need not be monopolized by statisticians, though, of course, technical statistical knowledge helps. Many
important ideas of technical statistics can be conveyed to the non-statistician without distortion or dilution.
Statistical interpretation depends not only on statistical ideas but also on ordinary clear thinking. Clear
thinking is not only indispensable in interpreting statistics but is often sufficient even in the absence of
specific statistical knowledge. For the statistician not only death and taxes but also statistical fallacies are
unavoidable. With skill, common sense, patience and above all objectivity, their frequency can be reduced
and their effects minimised. But eternal vigilance is the price of freedom from serious statistical blunders.”
We give below some illustrations regarding the mis-interpretation of statistical data.
1. “The number of car accidents committed in a city in a particular year by women drivers is 10 while
those committed by men drivers is 40. Hence women are safe drivers”. The statement is obviously wrong
since nothing is said about the total number of men and women drivers in the city in the given year. Some
valid conclusions can be drawn if we are given the proportion of the accidents committed by male and
female drivers.
2. “It has been found that the 25% of the surgical operations by a particular surgeon are successful. If
he is to operate on four persons on any day and three of the operations have proved unsuccessful, the fourth
must be a success.” The given conclusion is not true since statistical laws are probabilistic in nature and not
exact. The conclusion that if three operations on a particular day are unsuccessful, the fourth must be a
success, is not true. It may happen that the fourth operation is also unsuccessful. It may also happen that on
any day two or three or even all the four operations may be successful. The statement means that as the
number of operations becomes larger and larger, we should expect, on the average, 25% of the operations
to be successful.
3. A report : “The number of traffic accidents is lower in foggy weather than on clear weather days.
Hence it is safer to drive in fog.”
The statement again is obviously wrong. To arrive at any valid conclusions we must take into account
the difference between the rush of traffic under the two weather conditions and also the extra cautiousness
observed when driving in bad weather.
4. “80% of the people who drink alcohol die before attaining the age of 70 years. Hence drinking is
harmful for longevity of life.” This statement is also fallacious since no information is given about the
number of persons who do not drink alcohol and die before attaining the age of 70 years. In the absence of
the information about the proportion of such persons we cannot draw any valid conclusions.
5. Incomplete data usually leads us to fallacious conclusions. Let us consider the scores of two students
Ram and Shyam in three tests during a year.
1st test
2nd test
3rd test
Average Score
Ram’s Score
50%
60%
70%
60%
Shyam’s Score
70%
60%
50%
60%
If we are given the average score which is 60% in each case, we will conclude that the level of
intelligence of the two students at the end of the year is same. But this conclusion is false and misleading
since a careful study of the detailed marks over the three tests reveals that Ram has improved consistently
while Shyam has deteriorated consistently.
Remark. Numerous such examples can be constructed to illustrate the misuse of statistical methods
and this is all due to their unjudicious applications and interpretations for which the science of Statistics
cannot be blamed.
EXERCISE 1·1.
1. (a) Write a short essay on the origin and development of the science of Statistics.
(b) Give the names of some of the veterans in the development of Statistics, along with their contributions.
2. (a) Discuss the utility of Statistics to the state, the economist, the industrialist and the social worker.
(b) Define “Statistics” and discuss the importance of Statistics in a planned economy.
3. (a) Define the term “Statistics” and discuss its use in business and trade. Also point out its limitations.
[Punjab Univ. B.Com., April 1999]
1·14
(b) Define the term “Statistics” and discuss its functions and limitations.
(c) Explain the importance of Statistics with respect to business and industry.
BUSINESS STATISTICS
[C.S. (Foundation), June 2001]
[Delhi Univ. B.Com. (Pass), 2000]
4. Explain critically a few of the definitions of Statistics and state the one which you think to be the best.
5. (a) “Statistics is a method of decision-making in the face of uncertainty on the basis of numerical data and
calculated risks.” Explain with suitable illustrations.
(b) Discuss the scope of Statistics.
[Punjab Univ. B.Com., Oct. 1998]
6. Discuss briefly the importance of Statistics in the following disciplines :
(i) Economics
(i) Business and Management
(iii) Planning
(iv) Accountancy and Auditing
(v) Physical Sciences
(vi) Industry
(vii) Biology and Medical Sciences
(viii) Social Sciences
7. Comment briefly on the following statements :
(a) “Statistics is the science of human welfare.”
(b) “To a very striking degree our culture has become a statistical culture.”
(c) “Statistical thinking will one day be as necessary for efficient citizenship as the ability to read and write.”
8. (a) Comment briefly on the following statements :
(i) “Statistics can prove anything”.
(ii) “Statistics affects everybody and touches life at many points. It is both a science and an art”.
(b) “He who accepts statistics indiscriminately will often be duped unnecessarily, but he who distrusts statistics
indiscriminately will often be ignorant indiscriminately.” Comment on the above statement.
(c) “Sciences without statistics bear no fruit, statistics without sciences have no root.” Explain the above statement
with necessary comments.
9. Comment on the following statements illustrating your view point with suitable examples :
(a) “Knowledge of Statistics is like a knowledge of foreign language or of algebra. It may prove of use at any time
under any circumstances.” (Bowley)
(b) “Statistics is what statisticians do.”
(c) “There are lies, damned lies and Statistics – wicked in the order of their naming.”
(d) “By Statistics we mean aggregate of facts affected to a marked extent by multiplicity of causes, numerically
expressed, enumerated or estimated according to reasonable standards of accuracy, collected in a systematic manner for
a pre-determined purpose and placed in relation to each other’. (Horace Secrist)
(e) Statistics are the straws out of which I, like every other economist, have to make the bricks.” (Marshall)
(f) “Statistics conceals more than it reveals.”
[C.S. (Foundation), June 2002]
(g) “Statistics are like bikinis : they reveal what is interesting and conceal what is vital.”
10. (a) What do you understand by distrust of Statistics ? Is the science of Statistics to be blamed for it ?
(b) Write a critical note on the limitations and distrust of Statistics. Discuss the important causes of distrust and
show how Statistics could be made reliable.
(c) Define ‘Statistics’ and discuss its uses and limitations.
[Punjab Univ. B.Com., 1998]
(d) Discuss the use of Statistics in the fields of economics, trade and commerce. What are the limitations of
Statistics.
(e) Explain : “Distrust and misuse of Statistics.”
[C.S. (Foundation), June 2001]
(f) “Statistics widens the field of knowledge.” Elucidate the statement.
[C.S. (Foundation), June 2000]
11. (a) “Statistical methods are most dangerous tools in the hands of the inexperts.” Discuss and explain the
limitation of Statistics.
(b) “The science of Statistics, then, is a most useful servant but only of great value to those who understand its
proper use”–King.
Comment on the above statement and discuss the limitations of Statistics.
(c) “Statistics are like clay of which you can make a God or Devil, as you please.”
In the light of this statement, discuss the uses and limitations of Statistics.
(d) “All statistics are numerical statements but all numerical statements are not statistics.” Examine.
[C.S. (Foundation), Dec. 2000]
INTRODUCTION — MEANING & SCOPE
1·15
12. Comment on the following statements :
(a) “Statistics are like clay of which you make a God or Devil, as you please.”
(b) “Statistics is the science of estimates and probabilities.”
(c) “Statistics is the science of counting.”
(d) “Statistics should not be used as a blind man uses a lamp post for support, instead of for illumination.”
[C.S. (Foundation), Dec. 2001]
13. Comment on the following statistical statements, bringing out in details the fallacies, if any :
(i) “A survey revealed that the children of engineers, doctors and lawyers have high intelligence quotients (I.Q.). It
further revealed that the grandfathers of these children were also highly intelligent. Hence the inference is that
intelligence is hereditary.”
(ii) “The number of deaths in military in a recent war in a country was 15 out of 1,000 while the number of deaths
in the capital of the country during the same period was 22 per thousand. Hence it is safe to join military service than to
live in the capital city of the country.”
(iii) “The number of accidents taking place in the middle of the road is much less than the number of accidents
taking place on its sides. Hence it is safer to walk in the middle of the road.”
(iv) “The frequency of divorce for couples with the children is only about 12 of that for childless couples; therefore
producing children is an effective check on divorce.”
(v) “The increase in the price of a commodity was 20%. Then the price decreased by 15% and again increased by
10%. So the resultant increase in the price was 20 – 15 + 10 = 15%.”
(vi) “Nutritious Bread Company, a private manufacturing concern, charges a lower rate per loaf than that charged
by a Government of India Undertaking ‘Modern Bread.’ Thus private ownership is more efficient than public
ownership.”
(vii) According to the estimate of an economist, the per capita national income of India for 1931-32 was Rs. 65.
The National Income Committee estimated the corresponding figure for 1948-49 as Rs. 225. Hence in 1948-49, Indians
were nearly four times as prosperous as in 1931-32 ?
14. Point out the ambiguity or mistakes found in the following statements which are made on the basis of the facts
given :
(a) 80% of the people who die of cancer are found to be smokers and so it may be concluded that smoking causes
cancer.
(b) The gross profit to sales ratio of a company was 15% in the year 1994 and was 10% in 1995. Hence the stock
must have been undervalued.
(c) The average output in a factory was 2,500 units in January 1991 and 2,400 units in February 1991. So workers
were more efficient in January 1991.
(d) The rate of increase in the number of buffalloes in India is greater than that of the population. Hence the people
of India are now getting more milk per head.
15. Comment on the following :
(a) 50 boys and 50 girls took an examination. 30 boys and 40 girls got through the examination. Hence girls are
more intelligent than boys.
(b) The average monthly incomes in two cities of Hyderabad and Chennai were found to be Rs. 330. Hence, the
people of both the cities have the same standard of living.
(c) A tutorial college advertised that there was 100 per cent success of the candidates who took the coaching in
their institute. Hence the college has got good faculty.
16. Fill in the blanks :
(i) The word Statistics has been derived from the Latin word …… or the German word …… .
(ii) The word Statistics is used to convey different meanings in …… and …… sense.
(iii) Statistics is an …… and also a …… .
(iv) …… defined statistics as ‘numerical statement of facts’.
(v) In singular sense, Statistics means …… .
(vi) In plural sense, Statistics means …… .
(vii) Prof. Ya-Lun-Chou defined Statistics as ‘a method of ……’.
(viii) Bowley A.L. defined Statistics as …… counting.
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BUSINESS STATISTICS
(ix) Statistics are of …… help in human welfare.
(x) Prof. …… is the real giant in the development of the theory of Statistics.
(xi) “Statistics are the …… out of which I, like every other economist, have to make……” Alfred Marshall.
(xii) Two Indian statisticians who have made significant contribution in the development of statistics are …… and
…… .
Ans. (i) status, statistik; (ii) singular, plural; (iii) art, science; (iv) A.L. Bowley; (v) statistical methods used for
collecting analysing and drawing inferences from the numerical data; (vi) numerical set of data; (vii) decision making
in the face of uncertainty on the basis of numerical data and calculated risks. (viii) science of (ix) great (immense);
(x) R.A. Fisher; (xi) straws, bricks; (xii) P.C. Mahalanobis, C.R. Rao.
17. Indicate if the following statements are true (T) or false (F).
(i) The subject of statistics is a century old.
(ii) The word statistics seems to have been derived from Latin word status.
(iii) Statistics is of no use to humanity.
(iv) ‘To a very striking degree, our culture has become a statistical culture’.
(v) Statistics can prove anything.
Ans. (i) F; (ii) T; (iii) F; (iv) T; (v) F.
2
Collection of Data
2·1. INTRODUCTION
As pointed out in Chapter 1, Statistics are a set of numerical data. (See definitions of Secrist, Croxton
and Cowden, etc.). In fact only numerical data constitute Statistics. This means that the phenomenon under
study must be capable of quantitative measurement. Thus the raw material of Statistics always originates
from the operation of counting (enumeration) or measurement. For any statistical enquiry, whether it is in
business, economics or social sciences, the basic problem is to collect facts and figures relating to particular
phenomenon under study. The person who conducts the statistical enquiry i.e., counts or measures the
characteristics under study for further statistical analysis is known as investigator. Ideally, (though a costly
presumption), the investigator should be trained and efficient statistician. But in practice, this is not always
or even usually so. The persons from whom the information is collected are known as respondents and the
items on which the measurements are taken are called the statistical units. [For details see § 2·1·2]. The
process of counting or enumeration or measurement together with the systematic recording of results is
called the collection of statistical data. The entire structure of the statistical analysis for any enquiry is
based upon systematic collection of data.
On the face of it, it might appear that the collection of data is the first step for any statistical
investigation. But in a scientifically prepared (efficient and well-planned) statistical enquiry, the collection
of data is by no means the first step. Before we embark upon the collection of data for a given statistical
enquiry, it is imperative to examine carefully the following points which may be termed as preliminaries to
data collection :
(i) Objectives and scope of the enquiry.
(ii) Statistical units to be used.
(iii) Sources of information (data).
(iv) Method of data collection.
(v) Degree of accuracy aimed at in the final results.
(vi) Type of enquiry.
We shall discuss these points briefly in the following sections.
2·1·1. Objectives and Scope of the Enquiry. The first and foremost step in organising any statistical
enquiry is to define in clear and concrete terms the objectives of the enquiry. This is very essential for
determining the nature of the statistics (data) to be collected and also the statistical techniques to be
employed for the analysis of the data. The objectives of the enquiry would help in eliminating the collection
of irrelevant information which is never used subsequently and also reflect upon the uses to which such
information can be put. In the absence of the purpose of the enquiry being explicitly specified, we are
bound to collect irrelevant information and also omit some important information which will ultimately
lead to fallacious conclusions and wastage of resources.
Further, the scope of the enquiry will also have a great bearing upon the data to be collected and also
the techniques to be used for its collection and analysis. Scope of the enquiry relates to the coverage with
respect to the type of information, subject matter and geographical area. For instance, if we want to study
the cost of living index numbers, it must be specified if they relate to a particular city or state or whole of
2·2
BUSINESS STATISTICS
India. Further, the class of people (such as a low-income group, middle-income group, labour class, etc.),
for which they are intended should also be specified clearly. Thus, if the investigation is to be on a very
large scale, the sample method of enumeration and collection will have to be used. However, if the enquiry
is confined only to a small group, we may undertake 100% enumeration (census method). Thus if the scope
of the enquiry is very wide, it has to be of one nature and if the scope of enquiry is narrow, it has to be of a
totally different nature.
Thus the decision about the type of enquiry to be conducted depends largely on the objectives and
scope of the enquiry. However, the organisers of the enquiry should take care that these objectives and
scope are commensurate with the available resources in terms of money, manpower and time limit required
for the availability of the results of the enquiry.
2·1·2. Statistical Units to be Used. A well-defined and identifiable object or a group of objects with
which the measurements or counts in any statistical investigation are associated is called a statistical unit.
For example, in a socio-economic survey the unit may be an individual person, a family, a household or a
block of locality. A very important step before the collection of data begins is to define clearly the
statistical units on which the data are to be collected. In a number of situations the units are conventionally
fixed like the physical units of measurement such as metres, kilometres, kilograms, quintals, hours, days,
weeks, etc., which are well defined and do not need any elaboration or explanation. However in many
statistical investigations, particularly relating to socio-economic studies, arbitrary units are used which
must be clearly defined. This is imperative since in the absence of a clear-cut and precise definition of the
statistical units, serious errors in the data collection may be committed in the sense that we may collect
irrelevant data on the items, which should have, in fact, been excluded and omit data on certain items which
should have been included. This will ultimately lead to fallacious conclusions.
REQUISITES OF A STATISTICAL UNIT
The following points might serve as guidelines for deciding about the unit in any statistical enquiry.
1. It should be unambiguous. A statistical unit should be rigidly defined so that it does not lead to any
ambiguity in its interpretation. The units must cover the entire population and they should be distinct and
non-overlapping in the sense that every element of the population belongs to one and only one statistical
unit.
2. It should be specific. The statistical unit must be precise and specific leaving no chance to the
investigators. Quite often, in most of the socio-economic surveys the various concepts/characteristics can
be interpreted in different variant forms and accordingly the variable used to measure it may be defined in
several different ways. For example, in an enquiry relating to the wage level of workers in an industrial
concern the wages might by weekly wages, monthly wages or might refer to those of skilled labour only or
of day workers only or might include bonus payments also. Similarly prices in an enquiry might refer to
cost prices, selling prices, retail prices, wholesale prices or contract prices. Thus in a statistical enquiry it is
important to distinguish between the conventional and the arbitrary definitions of the
characteristics/variables, the former being the one prevalent in common use and shall always remain same
(fixed) for every enquiry while the latter is the one which is used in a specific sense and refers to the
working or operational definition which will keep on changing from one enquiry to another enquiry.
3. It should be stable. The unit selected should be stable over a long period of time and also w.r.t.
places i.e., there should not be significant fluctuations in the value of a unit at different intervals of time or
at different places because in the contrary case, the data collected at different times or places will not be
comparable and this would mar their utility to a great extent. The fluctuations in the value of money at
different times (due to inflation) or in the measurement of weights at different places (due to height above
sea level) might render the comparisons useless. Thus, the unit selected should imply, as far as possible, the
same characteristics at different times or at different places.
4. It should be appropriate to the enquiry. As already pointed out, the concept and definition of
arbitrary statistical units keep on changing from enquiry to enquiry. The unit selected must be relevant to
the given enquiry. Thus, for studying the changes in the general price level, the appropriate unit is the
wholesale prices while for constructing the cost of living indices (or consumer price indices) the
appropriate unit is the retail prices.
2·3
COLLECTION OF DATA
5. It should be uniform. It is essential that the unit adopted should be homogeneous (uniform)
throughout the investigation so that the measurements obtained are comparable. For example, in measuring
length if we use a yard on some occasions and metre on other occasions in an investigation, the
observations obtained would be confusing and misleading.
TYPES OF STATISTICAL UNITS
The statistical units may be broadly classified as follows :
(i) Units of collection.
(ii) Units of analysis and interpretation.
(i) Units of Collection. The units of collection may further be sub-divided into the following two
classes :
(a) Units of Enumeration. In any statistical enquiry, whether it is conducted by ‘sample’ method or
‘census’ method, unit of enumeration is the basic unit on which the observations are to be made and this
unit is to be decided in advance before conducting the enquiry keeping in view the objectives of the
enquiry. The unit of enumeration may be a person, a household, a family, a farm (in land experiments), a
shop, a livestock, a firm, etc. As has been pointed out earlier, this unit should be very clearly defined in
terms of shape, size, etc. For instance, for the construction of cost of living index number, the proper unit of
enumeration is household. It should be explained in clear terms whether a household consists of a family
comprising blood relations only or people taking food in a common kitchen or all the persons living in the
house or the persons enlisted in the ration card only. The concept of the household (to be used in the
enquiry) is to be decided in advance and explained clearly to the enumerators so that there are no essential
omissions or irrelevant inclusions.
(b) Units of Recording. The units of recording are the units in terms of which the data are recorded or
in other words they are the units of quantification. For instance, in the construction of cost of living index
number (consumer price index) the data to be collected from each household, among other things, include
the retail prices of various commodities together with the quantities consumed by the class of people for
whom the index is meant. The units of recording for quantity may be weight (in case of foodgrains), say, in
kilograms, quintals, tons, etc., in case of clothing the unit of recording may be metres; the prices may be
recorded in terms of rupees and so on.
Units of measurement (recording) may be simple or composite. The units which represent only one
condition without any qualification (adjective) are called simple units such as metre, rupee, ton, kilogram,
pound, bale of cloth, hour, week, year, etc. Such units are generally conventional and not at all difficult to
define. However, sometimes care has to be taken in their actual usage. For example, the bale of cloth must
be defined in terms of length, say, 20 metres, 50 metres or 100 metres. Similarly, in case of weight it should
be clearly specified whether it is net weight or gross weight.
A simple unit with some qualifying words is called a composite unit. A simple unit with only one
qualifying word is called a compound unit. Examples of such units are skilled worker, employed person,
ton–kilometre, kilowatt hour, man hours, retail prices, monthly wages, passenger kilometres. For instance
ton–kilometre means the number of tons multiplied by the number of kilometres carried; man hours implies
the total number of workers multiplied by the number of hours that each worker has put in and so on. If two
or more qualifying words are added to a simple unit, it is called a complex unit such as production per
machine hour, output per man hour and so on. Thus as compared to simple units, composite (compound and
complex) units are much more restrictive in scope and difficult to define. Such units should be defined
properly and clearly as they need explanation about the unit used and also about the qualifying words.
(ii) Units of Analysis and Interpretation. As the name implies, the units of analysis and interpretation
are those units in the form of which the statistical data are ultimately analysed and interpreted. It should be
decided whether the results would be expressed in absolute figures or relative figures. The units of analysis
and interpretation facilitate comparisons between different sets of data with respect to time, place or
environment (conditions). Generally, the units of analysis are rates, ratios and percentages, and coefficients.
Rates involve the comparison between two heterogeneous quantities i.e., when the numerator and
denominator are not of the same kind e.g., the mortality (death) rates, the fertility (birth) rates and so on.
Rates are usually expressed per thousand. For instance, the Crude Birth Rate (C.B.R.) is the ratio of total
2·4
BUSINESS STATISTICS
number of live births in the given region or locality during a given period to the total population of that
region or locality during the same period, multiplied by 1,000. Rate per unit is called coefficient. However,
ratios and percentages are used for comparing homogeneous quantities e.g., when the numerator and
denominator are of the same kind. For example, “the ratio of smokers to non-smokers in a particular
locality is 1 : 3” implies that 25% of the population are smokers.
From practical point of view for comparing data relating to different series, usually the unit of analysis
is one which gives relative figures which are pure numbers independent of units of measurement. For
instance, if we want to compare two series for variability (disperson) the appropriate unit of analysis is
Coefficient of Variation [See Chapter 6] and for comparing symmetry of two distributions, we study the
Coefficient of Skewness [See Chapter 7].
2·1·3. Sources of Information (Data). Having decided about the objectives and scope of the enquiry
and the statistical units to be used, the next problem is to decide about the sources from which the
information (data) can be obtained or collected. For any statistical enquiry, the investigator may collect the
data first hand or he may use the data from other published sources such as the publications of the
government/semi-government organisations, periodicals, magazines, newspapers, research journals, etc. If
the data are collected originally by the investigator for the given enquiry it is termed as primary data and if
he makes use of the data which had been earlier collected by some one else, it is termed as secondary data.
For example, the vital rates i.e., the rates of fertility and mortality in India prepared by the office of
Registrar-General of India, New Delhi, are primary data but if the same data are reproduced in the U.N.
Statistical Abstract (a publication of the United Nations Organisation), it becomes a secondary data.
Obviously the type of enquiry needed for primary data is bound to be of a totally different nature than the
type of enquiry needed for the use of secondary data. In case of primary data, the type of enquiry requires
laying down the definitions of the various terms and the statistical units used in the enquiry, keeping in
mind the objectives and scope of the enquiry. However, in the use of secondary data there are no such
problems since the data have already been collected under a given set of definitions of various terms and
units used. However, before using secondary data for statistical investigation under study, it must be
subjected to careful editing and scrutiny with respect to their reliability, suitability and adequacy. [For
details see § 2·6 in this Chapter]. For a given enquiry, the use of either primary or secondary data or both
may be made, depending upon the purpose and scope of the enquiry.
2·1·4. Method of Data Collection. The problem does not arise if secondary data are to be used.
However, if primary data are to be collected a decision has to be taken whether (i) census method or
(ii) sample technique, is to be used for data collection. In the census method, we resort to 100% inspection
of the population and enumerate each and every unit of the population. In the sample technique we inspect
or study only a selected representative and adequate fraction (finite subset) of the population and after
analysing the results of the sample data we draw conclusions about the characteristics of the population. In
some situations such as population being infinite or very large, census method fails. Moreover, it is not
practicable if the enumeration or testing of the units (objects) is destructive e.g., for testing the breaking
strength of chalk, testing the life of electric bulbs or tubes, testing of crackers and explosives, etc. Even, if
practicable, it may not be feasible from considerations of time and money. Thus a choice between the
sample method and census method is to be made depending upon the objectives and scope of the survey,
the limitations of resources in terms of time, money, manpower, etc., and the degree of accuracy desired. In
case of sample method the size of the sample and the technique of sampling like simple random sampling,
stratified random sampling, systematic sampling, etc., are to be decided. [For detailed discussion, see
Chapter 15 – Sampling Theory and Design of Sample Surveys].
2·1·5. Degree of Accuracy Aimed at in the Final Results. A decision regarding the degree of
accuracy or precision desired by the investigator in his estimates or results is essential before starting any
statistical enquiry. An idea about the precision aimed at is extremely helpful in deciding about the method
of data collection and the size of the sample (if the enquiry is to be on the basis of a sample study). The
information gained from any previous completed sample study on the subject in the form of precision
achieved for a given sample size may serve as a useful guide in this matter provided there is no
fundamental reason to change this empirical basis. In any statistical enquiry perfect accuracy in final results
is practically impossible to achieve because of the errors in measurement, collection of data, its analysis
and interpretation of the results. However, even if it were attainable, it is not generally desirable in terms of
COLLECTION OF DATA
2·5
time and money likely to be spent in attaining it and a reasonable degree of precision is enough to draw
valid inferences.
A decision regarding the precision of the results very much depends upon the objectives and scope of
the enquiry. For example, if we are measuring the length of cloth for shirting or pant, a difference of
centimetres is going to make substantial impact. But if we are measuring the distances between two places,
say, Delhi and Mumbai, a difference of few metres may be immaterial and while measuring the distance
between two distant places, say Delhi and New York (U.S.A.) a difference of few kilometres may be
immaterial. Likewise in measuring cereals (rice, wheat etc.) a difference of few grams may not matter at all
whereas in measuring gold even 1/15th or 1/20th part of a gram is going to make lot of difference. In the
words of Riggleman and Frisbee, “the necessary degree of accuracy in counting or measuring depends
upon the practical value of accuracy in relation to its cost.” However it should not be misunderstood to
imply that one should sacrifice accuracy to conduct the enquiry at low costs.
2·1·6. Types of Enquiry. Another important point one has to bear in mind before embarking upon the
process of collection of data is to decide about the type of enquiry. The statistical enquiries may be of
different types as outlined below :
(i) Official, Semi-official or Un-official.
(ii) Initial or Repetitive.
(iii) Confidential or Non-confidential.
(iv) Direct or Indirect.
(v) Regular or Ad-hoc.
(vi) Census or Sample.
(vii) Primary or Secondary.
(i) Official, Semi-official or Un-official Enquiry. A very important factor in the collection of data is
‘the sponsoring agency of the survey or enquiry’. If an enquiry is conducted by or on behalf of the central,
state or local governments it is termed as official enquiry. A semi-official enquiry is one that is conducted
by organisations enjoying government patronage like the Indian Council of Agricultural Research
(I.C.A.R.), New Delhi; Indian Agricultural Statistics Research Institute (I.A.S.R.I.), New Delhi; Indian
Statistical Institute (I.S.I.), Calcutta and New Delhi; and so on. An un-official enquiry is one which is
sponsored by private institutions like the F.I.C.C.I., trade unions, universities or the individuals. Obviously
the facilities available for each type of the above enquiries differ considerably. In an official enquiry, legal
or statutory compulsions can be exercised asking the public or respondents to furnish the requisite
information in time and that too at their own cost. In semi-official type enquiries also, the necessary
information may be obtained without much difficulty. However, in un-official enquiries the investigator is
faced with serious problems in getting information from the respondents. He can only pursuade and request
them for information. In such enquiries there is only moral obligation and no legal compulsion on the
respondents. Things are still worse if the enquiry is conducted by an individual who, at stages, has even to
beg for information. Moreover, there are lot of differences in the financial positions of these three
sponsoring agencies. Obviously, the state or central governments can afford to spend much more on an
enquiry as compared with private institutions, which in turn, can generally spend more than an individual.
Consequently, there is bound to be difference in the types of enquiries depending upon the sponsoring
agency and also its financial implications.
(ii) Initial or Repetitive Enquiry. As the name suggests, an initial or original enquiry is one which is
conducted for the first time while a repetitive enquiry is one which is carried on in continuation or
repetition of some previously conducted enquiry (enquiries). In conducting an original enquiry the entire
scheme of the plan starting with definitions of various terms, the units, the method of collection, etc., has to
be formulated afresh whereas in repetitive enquiry there is no such problem as such a plan already exists
and only the original enquiry is to be modified to suit the current situation and on the basis of the
experience gained in the past enquiry. However, for making valid conclusions in a repetitive enquiry, it
should be ascertained that there is not any material change in the definitions of various terms used in the
original enquiry.
(iii) Confidential or Non-confidential Enquiry. In a confidential enquiry, the information collected
and the results obtained are kept confidential and they are not made known to the public. The findings of
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BUSINESS STATISTICS
such enquiries are meant only for the personal records of the sponsoring agency. The enquiries conducted
by private organisations like trade unions, manufacturers’ associations, private business concerns, are
usually of confidential nature. On the other hand, the types of enquiries whose results are published and
made known to the general public are termed as non-confidential enquiries. Most of the enquiries
conducted by the state, private bodies or even individuals are of this type.
(iv) Direct or Indirect Enquiry. An enquiry is termed as direct if the phenomenon under study is
capable of quantitative measurement such as age, weight, income, prices, quantities consumed and so on.
However, if the phenomenon under study is of a qualitative nature which is not capable of quantitative
measurement like honesty, beauty, intelligence, etc., the corresponding enquiry is termed as indirect one. In
such an enquiry, the qualitative characteristic is converted into quantitative phenomenon by assigning
appropriate standard which may represent the given attribute (qualitative phenomenon) indirectly. For
example, the study of the attribute of intelligence may be made through the Intelligence Quotient (I.Q.)
score of a group of individuals in a given test.
(v) Regular or Ad-hoc Enquiry. If the enquiry is conducted periodically at equal intervals of time
(monthly, quarterly, yearly, etc.), it is said to be regular enquiry. For example, the census is conducted in
India periodically every 10 years. Similarly a number of enquiries are conducted by the Central Statistical
Organisation (C.S.O.) and their results are published periodically such as Monthly Abstract of Statistics,
Monthly Statistics of Production of Selected Industries of India, Statistical Abstract, India (Annually) ;
Statistical Pocket Book, India (Annually). On the other hand, if an enquiry is conducted as and when
necessary without any regularity or periodicity, it is termed as ad-hoc. For instance C.S.O. and N.S.S.O.
(National Sample Survey Organisation) conduct a number of ad-hoc enquiries.
(vi) and (vii). Enquiries of type (vi) viz., Census or Sample* and (vii) Primary or Secondary have
been discussed later in this chapter.
In any statistical enquiry, after deciding about the factors or problems enumerated above from § 2·1·1.
§ 2·1·6, we are now all set for the process of actual collection of data relating to the given enquiry. In the
following sections, we will discuss the methods of data collection.
2·2. PRIMARY AND SECONDARY DATA
After going through the preliminaries discussed in the above section, we come to the problem of data
collection. The most important factor in any statistical enquiry is that the original collection of data is
correct and proper. If there are inadequacies, shortcomings or pitfalls at the very source of the data, no
useful and valid conclusions can be drawn even after applying the best and sophisticated techniques of data
analysis and presentation of the results. In this context, it may be interesting to quote the remarks made by a
judge on Indian Statistics. “Cox, when you are bit older you will not quote Indian statistics with that
assurance. The governments are very keen on amassing statistics—they collect them, add them, raise them
to the nth power, take the cube root and prepare wonderful diagrams. But what you must never forget is that
every one of those figures comes in the first instance from the ‘Chowkidar’ (i.e., the village watchman)
who just puts down what he damn pleases.”**
It may be remarked that this quotation applies to India of very old days when no definite statistical set
up existed in India. Today in India, we have a fairly sound and systematic method of data collection on
almost all problems relating to various diversified fields such as economics, business, industry,
demography, social, physical and natural sciences. As already pointed out the data may be obtained from
the following two sources :
(i) The investigator or the organising agency may conduct the enquiry originally or
(ii) He may obtain the necessary data for his enquiry from some other sources (or agencies) who had
already collected the data on that subject.
The data which are originally collected by an investigator or agency for the first time for any statistical
investigation and used by them in the statistical analysis are termed as primary data. On the other hand, the
* For details see § 2·1·4 and Chapter 15.
** The earlierst use of this story seems to have been made in Sir Josiah Stamp, ‘Some Economic Factors in
Modern Life’, P.S. King and Son, London, 1929, p. 258-259.
COLLECTION OF DATA
2·7
data (published or unpublished) which have already been collected and processed by some agency or
person and taken over from there and used by any other agency for their statistical work are termed as
secondary data as far as second agency is concerned. The second agency if and when it publishes and files
such data becomes the secondary source to any one who later uses these data. In other words secondary
source is the agency who publishes or releases for use by others the data which was not originally collected
and processed by it.
It may be observed that the distinction between primary and secondary data is a matter of degree or
relativity only. The same set of data may be secondary in the hands of one and primary in the hands of
others. In general, the data are primary to the source who collects and processes them for the first time and
are secondary for all sources who later use such data. For instance, the data relating to mortality (death
rates) and fertility (birth rates) in India published by the Office of Registrar General of India, New Delhi are
primary whereas the same reproduced by the United Nations Organisation (U.N.O.) in its U.N. Statistical
Abstract become secondary in a far as later agency (U.N.O.) is concerned. For this data, the office of
Registrar General of India, is the primary source while U.N.O. is the secondary source. Likewise, the data
collected by C.S.O. and N.S.S.O. for various surveys are primary as far as these departments are concerned
but they become secondary if such data are used by other departments or organisations.
2·2·1. Choice Between Primary and Secondary Data. Obviously, there is lot of difference in the
method of collection of primary and secondary data. In the case of primary data which is to be collected
originally, the entire scheme of the plan starting with the definitions of various terms used, units to be
employed, type of enquiry to be conducted, extent of accuracy aimed at, etc., is to be formulated whereas
the collection of secondary data is in the form of mere compilation of the existing data. A proper choice
between the type of data (primary or secondary) needed for any particular statistical investigation is to be
made after taking into consideration the nature, objective and scope of the enquiry; the time and finances
(money) at the disposal of the agency; the degree of precision aimed at and the status of the agency
(whether government – state or central – or private institution or an individual).
Remarks 1. In using the secondary data it is best to obtain the data from the primary source as far as
possible. By doing so, we would at least save ourselves from the errors of transcription (if any) which
might have inadvertently crept in the secondary source. Moreover, the primary source will also provide us
with detailed discussion about the terminology used, statistical units employed, size of the sample and the
technique of sampling (if the sample method was used), methods of data collection and analysis of results
and we can ascertain ourselves if these suit our purpose.
2. It may be pointed out that today, in a large number of statistical enquiries secondary data are
generally used because fairly reliable published data on a large number of diverse fields are now available
in publications of the governments (state or centre), private organisations and research institutions,
international agencies, periodicals and magazines, etc. In fact primary data are collected only if there do not
exist any secondary data suited to the investigation under study. In some of the investigations both primary
as well as secondary data may be used.
3. Internal and External Data. Some statisticians differentiate between primary and secondary data in
the form of internal or external data. Internal data of an organisation (of business or economic concern or
firm) are those which are collected by the organisation from its own internal operations like production,
sales, profits, loans, imports and exports, capital employed, etc., and used by it for its own purposes. On the
other hand, external data are those which are obtained from the publications of some other agencies like
governments (central or state), international bodies, private research institutions, etc., for use by the given
organisation.
2·3. METHODS OF COLLECTING PRIMARY DATA
The methods commonly used for the collection of primary data are enumerated below :
(i) Direct personal investigation.
(ii) Indirect oral interviews.
(iii) Information received through local agencies.
(iv) Mailed questionnaire method.
(v) Schedules sent through enumerators.
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BUSINESS STATISTICS
2·3·1. Direct Personal Investigation. This method consists in the collection of data personally by the
investigator (organising agency) from the sources concerned. In other words, the investigator has to go to
the field personally for making enquiries and soliciting information from the informants or respondents.
This nature of investigation very much restricts the scope of the enquiry. Obviously this technique is suited
only if the enquiry is intensive rather than extensive. In other words, this method should be used only if the
investigation is generally local – confined to a single locality, region or area. Since such investigations
require the personal attention of the investigator, they are not suitable for extensive studies where the scope
of investigation is very wide. Obviously, the information gathered from such investigation is original in
nature.
Merits. (i) The first hand information obtained by the investigator himself is bound to be more reliable
and accurate since the investigator can extract the correct information by removing the doubts, if any, in the
minds of the respondents regarding certain questions. In case, the investigator suspects foul play on the part
of respondent(s) in supplying wrong information on certain items he can check it by some intelligent crossquestioning.
(ii) The data obtained from such investigation is generally reliable if the type of enquiry is intensive in
nature and if time and money do not pose any problems for the investigator.
(iii) When the audience is approached personally by the investigator, the response is likely to be more
encouraging.
(iv) Different persons have their own ideas, likes and dislikes and their opinions on some of the
questions may be coloured by their own prejudices and vision and as such some of them might react very
sharply to certain sensitive questions posed to them. The investigator, being on the spot, can handle such a
delicate situation creditably and effectively by his skill, intelligence and insight either by changing the topic
or if need be, by explaining to the respondent in polite words the objectives of the survey in detail.
(v) The investigator can extract proper information from the respondents by talking to them at their
educational level and if need be ask them questions in their language of communication and using local
connotations, if any, for the words used.
Demerits. (i) As already pointed out this type of investigation is restrictive in nature and is suited only
for intensive studies and not for extensive enquiries. This method is thus not suitable if the field of
investigation is too wide in terms of the number of persons to be interviewed or the area to be covered.
(ii) This type of investigation is handicapped due to lack of time, money and manpower (labour). It is
particularly time consuming since the informants can be approached only at their convenience and in case
of working class, this restricts the contact of the investigator with the informants only in the evenings or at
the week ends and consequently the investigation is to be spanned over a long period.
(iii) The greatest drawback of this enquiry is that it is absolutely subjective in nature. The success of
the investigation largely depends upon the intelligence, skill, tact, insight, diplomacy and courage of the
investigator. If the investigator lacks these qualities and is not properly trained, the results of the enquiry
cannot be taken as satisfactory or reliable. Moreover the personal biases, prejudices and whims of the
investigator may, in certain cases, adversely affect the findings of the enquiry.
(iv) Further, if the investigator is not intelligent, tactful or skillful enough to understand the
psychologies and customs of the interviewing audience, the results obtained from such an investigation will
not be reliable.
2·3·2. Indirect Oral Investigation. When the ‘direct personal investigation’ is not practicable either
because of the unwillingness or reluctance of the persons to furnish the requisite information or due to the
extensive nature of the enquiry or due to the fact that direct sources of information do not exist or are
unreliable, an indirect oral investigation is carried out. For example, if we want to solicit information on
certain social evils like if a person is addicted to drinking, gambling or smoking, etc., the person will be
reluctant to furnish correct information or he may give wrong information. The information on the
gambling, drinking or smoking habits of an individual can best be obtained by interviewing his personal
friends, relatives or neighbours who know him thoroughly well. In these types of enquiries factual data on
different problems are collected by interviewing persons who are directly or indirectly concerned with the
subject matter of the enquiry and who are in possession of the requisite information. The method consists in
collection of the data through enumerators appointed for this purpose. A small list of questions pertaining to
COLLECTION OF DATA
2·9
the subject matter of the enquiry is prepared. These questions are then put to the persons, known as
witnesses or informants, who are in possession of such information and their replies are recorded. Such a
procedure for the collection of factual data on different problems is usually adopted by the Enquiry
Committees or Commissions appointed by the government—State or Central.
Merits. (i) Since the enumerators contact the informants personally, as discussed in the first method,
they can exercise their intelligence, skill, tact, etc., to extract correct and relevant information by cross
examination of the informants, if necessary.
(ii) As compared with the method of “direct personal investigation”, this method is less expensive and
requires less time for conducting the enquiry.
(iii) If necessary, the expert views and suggestions of the specialists on the given problem can be
obtained in order to formulate and conduct the enquiry more effectively and efficiently.
Demerits. (i) Due to lack of direct supervision and personal touch the investigator (sponsoring agency)
has to rely entirely on the information supplied by the enumerators. The success of the method lies in the
intelligence, skill, insight and efficiency of the enumerators and also on the fact that they are honest persons
with high integrity and without any selfish motives. It should be ascertained that the enumerators are
properly trained and tactful enough to elicit proper and correct response from the informants. Moreover, it
should be seen that the personal biases due to the prejudices, and whims of the enumerators do not enter or
at least they are minimised.
(ii) The accuracy of the data collected and the inferences drawn, depend to a large extent on the nature
and quality of the witnesses from whom the information is obtained. A wrong and improper choice of the
witnesses will give biased results which may adversely affect the findings of the enquiry. It is, therefore,
imperative :
(a) To ascertain the reliability and integrity of the persons (witnesses) selected for interrogation. In
other words, it should be ascertained that the witnesses are unbiased persons without any selfish motives
and that they are not prejudiced in favour of or against a particular view point.
(b) That the findings of the enquiry are not based on the information supplied by a single person alone.
Rather, a sufficient number of persons should be interviewed to find out the real position.
(c) That the witnesses really possess the knowledge about the problem under study i.e., they are aware
of the full factors of the problem under investigation and are in a position to give a clear, detailed and
correct account of the problem.
(d) That a proper allowance about the pessimism or optimism of the witnesses depending upon their
inherent psychology should be made.
2·3·3. Information Received Through Local Agencies. In this method the information is not
collected formally by the investigator or the enumerators. This method consists in the appointment of local
agents (commonly called correspondents) by the investigator in different parts of field of enquiry. These
correspondents or agencies in different regions collect the information according to their own ways,
fashions, likings and decisions and then submit their reports periodically to the central or head office where
the data are processed for final analysis. This technique of data collection is usually employed by
newspaper or periodical agencies who require information in different fields like sports, riots, strikes,
accidents, economic trend, business stock and share market, policies and so on. This method is also used by
the various departments of the government (state or central) where the information is desired periodically
(at regular intervals of time) from a wide area. This method is particularly useful in obtaining the estimates
of agricultural crops which may be submitted to the government by the village school teachers. A more
refined and sophisticated way of the use of this technique is the registration method in which any event,
say, birth, death, incidence of disease, etc., is to be reported to the appropriate authority appointed by the
government like Sarpanch or Patwari in the village; or Block Development Officers (B.D.O.’s), civil
hospitals or the health departments in the district headquarters, etc., as and when or immediately after it
occurs. Vital statistics i.e., the data relating to mortality (deaths) and fertility (births) are usually collected
in India through the registration technique.
Merits. This method works out to be very cheap and economical for extensive investigations
particularly if the data are obtained through part-time correspondents or agents. Moreover, the required
information can be obtained expeditiously since only rough estimates are required.
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BUSINESS STATISTICS
Demerits. Since the different correspondents collect the information in their own fashion and style, the
results are bound to be biased due to the personal prejudices and whims of the correspondents in different
fields of the enquiry and consequently the data so obtained will not be very reliable. Hence, this technique
of data collection is suited if the purpose of investigation is to obtain rough and approximate estimates only
and where a high degree of accuracy is not desired.
In particular, the registration method suffers from the drawback that many persons do not report and
thus neglect to register. This usually results in under-estimation. For an effective and efficient system of
registration there should be legal compulsions for registration of events and also there should be sanctions
for the enforcement of the obligation.
2·3·4. Mailed Questionnaire Method. This method consists in preparing a questionnaire (a list of
questions relating to the field of enquiry and providing space for the answers to be filled by the
respondents) which is mailed to the respondents with a request for quick response within the specified time.
A very polite covering note, explaining in detail the aims and objectives of collecting the information and
also the operational definitions of various terms and concepts used in the questionnaire is attached.
Respondents are also requested to extend their full co-operation by furnishing the correct replies and
returning the questionnaire duly filled in time. Respondents are also taken into confidence by ensuring them
that the information supplied by them in the questionnaire will be kept strictly confidential and secret. In
order to ensure quick and better response the return postage expenses are usually borne by the investigator
by sending a self-addressed stamped envelope. This method is usually used by the research workers, private
individuals, non-official agencies and sometimes even by government (central or state).
In this method, the questionnaire is the only media of communication between the investigator and the
respondents. Consequently, the most important factor for the success of the ‘mailed questionnaire method’,
is the skill, efficiency, care and the wisdom with which the questionnaire is framed. The questions asked
should be clear, brief, corroborative, non-offending, courteous in tone, unambiguous and to the point so that
not much scope of guessing is left on the part of the respondent. Moreover, while framing the questions the
knowledge, understanding and the general educational level of the respondents should be taken into
consideration.
Remark. “Drafting or framing the questionnaire”, is of paramount importance and is discussed in
detail in § 2·4 after ‘Schedules sent through enumerators’.
Merits. (i) Of all the methods of collecting information, the ‘mailed questionnaire method’ is by far the
most economical method in terms of time, money and manpower (labour) provided the respondents supply
the information in time.
(ii) This method is used for extensive enquiries covering a very wide area.
(iii) Errors due to the personal biases of the investigators or enumerators are completely eliminated as
the information is supplied directly by the person concerned in his own handwriting. The information so
obtained is original and much more authentic.
Demerits. (i) The most serious drawback of this method is that it can be used effectively with
advantage only if the audience (people) is (are) educated (literate) and can understand the questions well
and reply them in their own handwriting. Obviously, this method is not practicable if the people are
illiterate. Even if they are educated, there may be a number of persons who are not interested in the
particular enquiry being conducted and as such they adopt an attitude of indifference towards the enquiry
which results in their questionnaires finding place in the waste paper baskets. In the case of those who
return the questionnaires after filling, a number of them supply haphazard, vague, incomplete and
unintelligible information which does not serve much purpose. Thus, this method generally suffers from the
high degree (i.e., very large proportion) of non-response and consequently the results based on the
information supplied by a very small proportion of the selected individuals cannot be regarded as reliable.
(ii) Quite often people might suppress correct information and furnish wrong replies. We cannot verify
the accuracy and reliability of the information received. In general, this method also suffers from the low
degree of reliability of the information supplied by the respondents.
(iii) Another limitation of this method is that at times, informants are not willing to give written
information in their own handwriting on certain personal questions like income, property, personal habits
and so on.
COLLECTION OF DATA
2·11
(iv) Since the questionnaires are filled by the respondents personally, there is no scope for asking
supplementary questions for cross checking of the information supplied by them. Moreover, the doubts in
the minds of the informants, if any, on certain questions cannot be dispelled.
2·3·5. Schedules Sent Through Enumerators. Before discussing this method it is desirable to make a
distinction between a questionnaire and a schedule. As already explained, questionnaire in a list of
questions which are answered by the respondent himself in this own handwriting while schedule is the
device of obtaining answers to the questions in a form which is filled by the interviewers or enumerators
(the field agents who put these questions) in a face to face situation with the respondents. The most widely
used method of collection of primary data is the ‘schedules sent through the enumerators’. This is so
because this method is free from certain shortcomings inherent in the earlier methods discussed so far. In
this method the enumerators go to the respondents personally with the schedule (list of questions), ask them
the questions there in and record their replies. This method is generally used by big business houses, large
public enterprises and research institutions like National Council of Applied Economic Research (NCAER),
Federation of Indian Chambers of Commerce and Industries (FICCI) and so on and even by the
governments – state or central – for certain projects and investigations where high degree of response is
desired. Population census, all over the world is conducted by this technique.
Merits. (i) The enumerators can explain in detail the objectives and aims of the enquiry to the
informants and impress upon them the need and utility of furnishing the correct information. Being on the
spot, the enumerators can dispel the doubts, if any, of certain people to certain questions by explaining to
them the implications of certain definitions and concepts used in the questionnaire.
(ii) This technique is very useful in extensive enquiries and generally yields fairly dependable and
reliable results due to the fact that the information is recorded by highly trained and educated enumerators.
Moreover, since the enumerators personally call on the respondents to obtain information there is very little
non-response which occurs if it is not possible to contact the respondents even after repeated calls or if the
respondent is unwilling to furnish the requisite information. Thus, this method removes both the drawbacks
of the ‘mailed questionnaire method’, viz., very large proportion of non-response and fairly low degree of
reliability of the information.
(iii) Unlike the ‘mailed questionnaire method’, this technique can be used with advantage even if the
respondents are illiterate.
(iv) As already pointed out in the ‘direct personal investigation’, due to personal likes and dislikes,
different people react differently to different questions and as such some people might react very sharply to
certain sensitive and personal questions. In that case the enumerators, by their tact, skill, wisdom and
calibre can handle the situation very effectively by changing the topic of discussion, if need be. Moreover,
the enumerators can effectively check the accuracy of the information supplied by some intelligent crossquestioning by asking some supplementary questions.
Demerits. (i) It is fairly expensive method since the team of enumerators is to be paid for their services
and as such can be used by only those bodies or institutions which are financially sound.
(ii) It is also more time consuming as compared with the ‘mailed questionnaire method’.
(iii) The success of the method largely depends upon the efficiency and skill of the enumerators who
collect the information. Thus the choice of enumerators is of paramount importance. The enumerators have
to be trained properly in the art of collecting correct information by their intelligence, insight, patience and
perseverence, diplomacy and courage. They should clearly understand the aims and objectives of the
enquiry and also the implications of the various terms, definitions and concepts used in the questionnaire.
They should be provided with adequate guidelines so that their personal biases do not enter the final results
of the enquiry. They should be honest persons with high integrity and should not have any personal axe to
grind. They should be well versed in the local language, customs and traditions. If the enumerators are
biased they may suppress or even twist the information supplied by the respondents. Inefficiency on the
part of the enumerators coupled with personal biases due to their prejudices and whims will lead to false
conclusions and may even adversely affect the results of the enquiry.
(iv) Due to inherent variation in the individual personalities of the enumerators there is bound to be
variation, though not so obvious, in the information recorded by different enumerators. An attempt should
be made to minimise this variation.
2·12
BUSINESS STATISTICS
(v) The success of this method also lies to a great extent on the efficiency and wisdom with which the
schedule is prepared or drafted. If the schedule is framed haphazardly and incompetently, the enumerators
will find it very difficult to get the complete and correct desired information from the respondents.
Remarks. 1. In the last two methods viz., ‘mailed questionnaire method’ and the ‘schedules sent
through enumerators’, it is desirable to scrutinise the questionnaires or schedules duly filled in for detecting
any apparent inconsistency in the information supplied by the respondents or recorded by the enumerators.
2. If resources (time, money and manpower) permit, two sets of enumerators may be used for recording
information for the enquiry under investigation and their findings may be compared. This will, incidentally,
provide a check on the honesty and integrity of the enumerators and will also reflect upon personal bias due
to the prejudices and whims of the individual personalities (of the enumerators). However, this technique is
not practicable in the case of interviewing individuals, who might get irritated, annoyed or confused when
approached for the second time.
2·4. DRAFTING OR FRAMING THE QUESTIONNAIRE
As has been pointed out earlier, the questionnaire is the only media of communication between the
investigator and the respondents and as such the questionnaire should be designed or drafted with utmost
care and caution so that all the relevant and essential information for the enquiry may be collected without
any difficulty, ambiguity and vagueness. Drafting of a good questionnaire is a highly specialised job and
requires great care, skill, wisdom, efficiency and experience. No hard and fast rules can be laid down for
designing or framing a questionnaire. However, in this connection, the following general points may be
borne in mind :
1. The size of the questionnaire should be as small as possible. The number of questions should be
restricted to the minimum, keeping in view the nature, objectives and scope of the enquiry. In other words,
the questionnaire should be concise and should contain only those questions which would furnish all the
necessary information relevant for the purpose. Respondents’ time should not be wasted by asking
irrelevant and unimportant questions. A large number of questions would involve more work for the
investigator and thus result in delay on his part in collecting and submitting the information. These may, in
addition, also unnecessarily annoy or tire the respondents. A reasonable questionnaire should contain from
15 to 20-25 questions. If a still larger number of questions is a must in any enquiry, then the questionnaire
should be divided into various sections or parts.
2. The questions should be clear, brief, unambiguous, non-offending, courteous in tone, corroborative
in nature and to the point so that not much scope of guessing is left on the part of the respondents.
3. The questions should be arranged in a natural logical sequence. For example, to find if a person
owns a refrigerator the logical order of questions would be : “Do you own a refrigerator”? When did you
buy it ? What is its make ? How much did it cost you ? Is its performance satisfactory ? Have you ever got
it serviced ? The logical arrangement of questions in addition to facilitating tabulation work, would leave
no chance for omissions or duplication.
4. The usage of vague and ‘multiple meaning’ words should be avoided. The vague works like good,
bad, efficient, sufficient, prosperity, rarely, frequently, reasonable, poor, rich, etc., should not be used since
these may be interpreted differently by different persons and as such might give unreliable and misleading
information. Similarly the use of words with multiple meanings like price, assets, capital, income,
household, democracy, socialism, etc., should not be used unless a clarification to these terms is given in
the questionnaire.
5. Questions should be so designed that they are readily comprehensible and easy to answer for the
respondents. They should not be tedious nor should they tax the respondents’ memory. Further, questions
involving mathematical calculations like percentages, ratios, etc., should not be asked.
6. Questions of a sensitive and personal nature should be avoided. Questions like ‘How much money
you owe to private parties ?’ or ‘Do you clean your utensils yourself ?’ which might hurt the sentiments,
pride or prestige of an individual should not be asked, as far as possible. It is also advisable to avoid
questions on which the respondent may be reluctant or unwilling to furnish information. For example, the
questions pertaining to income, savings, habits, addiction to social evils, age (particularly, in case of ladies),
etc., should be asked very tactfully.
2·13
COLLECTION OF DATA
7. Typed of Questions. Under this head, the questions in the questionnaire may be broadly classified as
follows :
(a ) Shut Questions. In such questions possible answers are suggested by the framers of the
questionnaire and the respondent is required to tick one of them. Shut questions can further be sub-divided
into the following forms :
(i) Simple Alternate Questions. In such questions, the respondent has to choose between two clear cut
alternatives like ‘Yes or No’ ; ‘Right or Wrong’ ; ‘Either, Or’ and so on. For instance, Do you own a
refrigerator ?—Yes or No. Such questions are also called dichotomous questions. This technique can be
applied with elegance to situations where two clear cut alternatives exist.
(ii) Multiple Choice Questions. Quite often, it is not possible to define a clear cut alternative and
accordingly in such a situation either the first method (Alternate Questions) is not used or additional
answers between Yes and No like Do not know, No opinion, Occasionally, Casually, Seldom, etc., are
added. For instance to find if a person smokes or drinks, the following multiple choice answers may be
used :
Do you smoke ?
Yes (Regularly)
■
No (Never)
■
Occasionally
■
Seldom
■
Similarly, to get information regarding the mode of cooking in a household, the following multiple
choice answers may be suggested.
Which of the following modes of cooking you use ?
Gas
■
Coal (Coke)
■
Power (Electricity)
■
Wood
■
Stove (Kerosene)
■
As another illustration, to find what conveyance an individual uses to go from his house to the place of
his duty, the following question with multiple answers may be framed :
How do you go to your place of duty ?
By bus
■
By three wheeler scooter ■
By your own cycle
■
By taxi
■
By your own scooter/Motor cycle ■
On foot
■
By your own car
■
Any other
■
Multiple choice questions are very easy and convenient for the respondents to answer. Such questions
save time and also facilitate tabulation. This method should be used if only a selected few alternative
answers exist to a particular question. Sometimes, a last alternative under the category ‘Others’ or ‘Any
other’ may be added. However, multiple answer questions cannot be used with advantage if it is possible to
construct a fairly large number of alternative answers of relatively equal importance to a given question.
(b) Open Questions. Open questions are those in which no alternative answers are suggested and the
respondents are at liberty to express their frank and independent opinions on the problem in their own
words. For instance, ‘What are the drawbacks in our examination system’ ? ; ‘What solution do you suggest
to the housing problem in Delhi’ ? ; ‘Which programme in the Delhi TV do you like best’ ? ; are some of
the open questions. Since the views of the respondents in the open questions might differ widely, it is very
difficult to tabulate the diverse opinions and responses.
Remark. Sometimes a combination of both shut questions and open questions might be used. For
instance an open question : ‘When did you buy the car’ ?, may be followed by a multiple choice question as
to whether its performance is (i) extremely good, (ii) satisfactory, (iii) poor, (iv) needs improvement.
8. Leading questions should be avoided. For example, the question ‘Why do you use a particular brand
of blades, say, Erasmic blades’ should preferably be framed into two questions.
(i) Which blade do you use ?
2·14
(ii) Why do you prefer it ?
Gives a smooth shave
Gives more shaves
Price is less (Cheaper)
BUSINESS STATISTICS
■
■
■
Readily available in the market
Any other
■
■
9. Cross Checks. The questionnaire should be so designed as to provide internal checks on the accuracy
of the information supplied by the respondents by including some connected questions at least with respect
to matters which are fundamental to the enquiry. For example in a social survey for finding the age of the
mother the question ‘What is your age’ ?, can be supplemented by additional questions ‘What is your date
of birth’ or ‘What is the age of your eldest child ?’ Similarly, the question, ‘Age at marriage’ can be
supplemented by the question ‘The age of the first child’.
10. Pre-testing the Questionnaire. From practical point of view it is desirable to try out the
questionnaire on a small scale (i.e., on a small cross-section of the population for which the enquiry is
intended) before using it for the given enquiry on a large scale. This testing on a small scale (called pretest) has been found to be extremely useful in practice. The given questionnaire can be improved or
modified in the light of the drawbacks, shortcomings and problems faced by the investigator in the pre-test.
Pre-testing also helps to decide upon the effective methods of asking questions for soliciting the requisite
information.
11. A Covering Letter. A covering letter from the organisers of the enquiry should be enclosed along
with the questionnaire for the following purposes :
(i) It should clearly explain in brief the objectives and scope of the survey to evoke the interest of the
respondents and impress upon them to render their full co-operation by returning the schedule/questionnaire
duly filled in within the specified period.
(ii) It should contain a note regarding the operational definitions to the various terms and the concepts
used in the questionnaire; units of measurements to be used and the degree of accuracy aimed at.
(iii) It should take the respondents in confidence and ensure them that the information furnished by
them will be kept completely secret and they will not be harassed in any way later.
(iv) In the case of mailed questionnaire method a self-addressed stamped envelope should be enclosed
for enabling the respondents to return the questionnaire after completing it.
(v) To ensure quick and better response the respondents may be offered awards/incentives in the form
of free gifts, coupons, etc.
(vi) A copy of the survey report may be promised to the interested respondents.
12. Mode of tabulation and analysis viz., hand operated, machine tabulation or computerisation should
also be kept in mind while designing the questionnaire.
13. Lastly, the questionnaire should be made attractive by proper layout and appealing get up.
We give below two specimen questionnaires for illustration.
MODEL I
Questionnaire for Collecting Information (Covering Production, Employment etc.) Relating to an
Industrial concern
1. Name of the concern . . . . . .
2. (a) Name of the Proprietor/Managing Director . . . . . .
(b) Qualifications :
(i) Academic . . . . . .
(ii) Technical/Professional . . . . . .
3. (a) Location
(b) Telephone Number
(c) e-mail Address
(i) Factory . . . . . .
(i) Factory . . . . . .
(i) Factory . . . . . .
(ii) Office . . . . . .
(ii) Office . . . . . .
(ii) Office . . . . . .
4. Factory Registration Number (with date) . . . . . .
5. Total Capital employed/Assets (approximately) . . . . . .
6. Number of Shifts . . . . . .
2·15
COLLECTION OF DATA
7. Whether the machinery used is indigenous ?
Yes ■
No ■
Other ■
8. What is the approximate value of the imported machinery ? . . . . . .
9. Whether the raw material used is available in domestic market ?
Yes ■
No ■
10. From which country is the raw material imported ? . . . . . .
11. What is the approximate annual consumption of the raw material ? . . . . . .
12. Expenditure in foreign currency :
(i) Foreign Travel . . . . . . ; (ii) Technical know-how . . . . . . ; (iii) Material and goods . . . . .
13. Employment (Pay-roll) :
S. No.
Categories
Working
No. employed Salaries paid in
hours per
’000 Rs.
week
2001
2002
2001
2002
(i)
Management
(ii)
Supervisory/Technical Personnel
(iii)
Skilled workers
(iv)
Unskilled workers
(v)
Non-technical office staff
14. Production :
S. No.
Items (Production)
Installed Capacity
Actual Production
2001
2002
15. Market
2001
2002
(a) Gross value of sales (’000 Rs.)
......
......
(b) Who are :
(i) Immediate purchasers . . . . . .
(ii) End users . . . . . .
(c) The extent of market is
Local
■
National
■
International
■
(d) If the extent of the market is international
(i) What are the approximate foreign exchange earnings (annually) ?
(ii) Which countries are the chief importers of the product ?
(e) Total Sales (’000 Rs.) :
National market : . . . . . .
;
International market : . . . . . .
(f) Are the present conditions of the market satisfactory/not satisfactory/poor ? . . . . . .
16. Financial Highlights
(Rupees ’000s)
Items
2001
2002
Sales
Profits
Dividends
Capital expenditure
Fixed assets
Shareholders’ funds
2·16
BUSINESS STATISTICS
MODEL II
We
collect :
(i)
(ii)
(iii)
1.
2.
3.
6.
7.
8.
9.
11.
13.
15.
16.
give below the 1971 Census – Individual Slip which was used for a general purpose survey to
Social and cultural data like nationality, religion, literacy, mother tongue, etc.;
Exhaustive economic data like occupation, industry, class of worker and activity, if not working ;
Demographic data like relation to the head of the house, sex, age, marital status, birth place, births
and deaths and the fertility of women to assess in particular the performance of the family
planning programme.
1971 CENSUS – INDIVIDUAL SLIP
Name...........................................................
Relationship to the head of the family.....................................................
Sex............................... ; 4. Age............................ ; 5. Marital status................................
For currently married women only :
(a) Age at marriage...........
(b) Any child born in the last one year..............
Birth place :
(i) Place of birth...............
(ii) Rural or urban..................
(iii) District........................
(iv) State/country....................
Last Residence :
(i) Place of last residence..............
(ii) Rural/urban.................
(iii) District...................
(iv) State/country...............
Duration of present residence.............
10. Religion........................
Scheduled Caste or Tribe..............
12. Literacy.........................
Educational level..............
14. Mother tongue.................
Other languages, if any................
Main activity :
(a) Broad category : (i) Worker (C, AL, HHI, OW)* (ii) Non-worker (H, ST, R, D.B.I.O.)**
(b) Place of work (Name of Village/Town) . . . . . .
(c) Name of establishment…
(d) Name of Industry, Trade, Profession or Service . . . . . .
(e) Description of work . . . . . . (f) Class of worker . . . . . .
17. Secondary work :
(a) Broad category (C, AL, HHI, OW) (b) Place of work . . . . . .
(c) Name of establishment . . . . . .
(d) Nature of Industry, Trade, Profession or Service . . .
(e) Description of work . . . . . .
(f) Class of worker . . . . . .
2·5. SOURCES OF SECONDARY DATA
The chief sources of secondary data may be broadly classified into the following two groups :
(i) Published sources.
(ii) Unpublished sources.
2·5·1. Published Sources. There are a number of national (government, semi-government and private)
organisations and also international agencies which collect statistical data relating to business, trade, labour,
prices, consumption, production, industries, agriculture, income, currency and exchange, health, population
*C
AL
HHI
OW
: Cultivator
: Agriculture Labour
: House Hold Industries
: Other Works
**H
ST
R
DBIO
: Household Duties
: Student
: Retired person or Renteer
: Dependent, Beggar, Institutions, Others
COLLECTION OF DATA
2·17
and a number of socio-economic phenomena and publish their findings in statistical reports on a regular
basis (monthly, quarterly, annually, ad-hoc). These publications of the various organisations serve as a very
powerful source of secondary data. We give below a brief summary of these sources.
1. Official Publications of Central Government. The following are various government organisations
along with the year of their establishment [given in bracket ( )] which collect, compile and publish
statistical data on a number of topics of current interest – prices, wages, population, production and
consumption, labour, trade, army, etc.
(1) Office of the Registrar General and Census Commissioner of India, New Delhi (1949).***
(2) Directorate-General of Commercial Intelligence and Statistics – Ministry of Commerce (1895).
(3) Labour Bureau – Ministry of Labour (1946).
(4) Directorate of Economics and Statistics – Ministry of Agriculture and Irrigation (1948).
(5) The Indian Army Statistical Organisation (I.A.S.O.) – Ministry of Defence (1947).
(6) National Sample Survey Organisation (N.S.S.O.), Department of Statistics, Ministry of Planning
(1950).****
(7) Central Statistical Organisation (C.S.O.) – Department of Statistics, Ministry of Planning (1951).
Some of the main publications of the above government agencies are :
(a) Monthly Abstract of Statistics ; Monthly Statistics of Production of Selected Industries in India ;
Statistical Pocket Book, India ; Annual Survey of Industries – General Review ; Sample Surveys of Current
Interest in India (all published annually) ; Statistical Systems of India ; National Income Statistics –
Estimates of Savings in India (1960-61 to 1965-66) ; National Income Statistics – Estimates of Capital
Formation in India (1960-61 to 1965-66) (Ad-hoc publications) ; all published by the Central Statistical
Organisation (C.S.O.), New Delhi.
(b) Census data in various census reports ; Vital Statistics of India (Annual), Indian Population Bulletin
(Biennial) – all published by Registrar-General of India (R.G.I.).
(c) Various statistical reports on phenomenon relating to socio-economic and demographic conditions,
prices, area and yield of different crops, as a result of the various surveys conducted in different rounds by
National Sample Survey Organisation (N.S.S.O.).
In addition to the above organisations a number of departments in the State and Central Governments
like Income Tax Department, Directorate General of Supplies and Disposals, Railways, Post and
Telegraphs, Central Board of Revenues, Textile Commissioner’s Office, Central Excise Commissioner’s
Office, Iron and Steel Controller’s Office and so on, publish statistical reports on current problems and the
information supplied by them is, in general, more authentic and reliable than that obtained from other
sources on the same subject.
2. Publications of Semi-Government Statistical Organisations. Very useful information is provided by
the publications of the semi-government statistical organisations some of which are enumerated below :
(i) Statistics department of the Reserve Bank of India (Mumbai), which brings out an Annual Report of
the Bank, Currency and Finance ; Reserve Bank of India Bulletin (monthly) and various monthly and
quarterly reports.
(ii) Economic department of Reserve Bank of India ;
(iii) The Institute of Economic Growth, Delhi.
(iv) Gokhale Institute of Politics and Economics, Poona ; (v) The Institute of Foreign Trade, New Delhi.
Moreover, the statistical material published by the institutions like Municipal and District Boards,
Corporations, Block and Panchayat Samitis on Vital Statistics (births and deaths), health, sanitation and
other related subjects provides a fairly reliable and useful information.
3. Publications of Research Institutions. Individual research scholars, the different departments in the
various universities of India and various research organisations and institutes like Indian Statistical Institute
*** In India census has been carried out every ten years since 1881. Prior to the establishment of this organisation,
the census was conducted by a temporary cell in the Ministry of Home Affairs.
**** National Sample Survey (N.S.S.) was set up in 1950 in Ministry of Finance. In 1957, N.S.S. was transferred
to the Cabinet Secretariate and named National Sample Survey Organisation (N.S.S.O.).
2·18
BUSINESS STATISTICS
(I.S.I.), Kolkata and Delhi ; Indian Council of Agricultural Research (I.C.A.R.), New Delhi ; Indian
Agricultural Statistics Research Institute (I.A.S.R.I.), New Delhi ; National Council of Educational
Research and Training (N.C.E.R.T.), New Delhi ; National Council of Applied Economic Research, New
Delhi ; The Institute of Applied Man Power Research, New Delhi ; The Institute of Labour Research,
Mumbai ; Indian Standards Institute, New Delhi ; and so on publish the findings of their research
programmes in the form of research papers, or monographs or journals which are a constant source of
secondary data on the subjects concerned.
4. Publications of Commercial and Financial Institutions. A number of private commercial and trade
institutions like Federation of Indian Chamber of Commerce and Industries (FICCI), Institute of Chartered
Accountants of India, Trade Unions, Stock Exchanges, Bank Bodies, Co-operative Societies, etc., publish
reports and statistical material on current economic, business and other phenomena.
5. Reports of Various Committees and Commissions appointed by the Government. The report of the
survey and enquiry commissions and committees of the Central and State Governments to find their expert
views on some important matters relating to economic and social phenomena like wages, dearness
allowance, prices, national income, taxation, land, education, etc., are invaluable source of secondary
information. For instance Simon-Kuznet Committee report on National Income in India, Wanchoo
Commission report on Taxation, Kothari Commission report on Educational Reforms, Pay Commissions
Reports, Land Reforms Committee report, Gupta Commission report on Maruti Affairs, etc., are invaluable
sources of secondary data.
6. Newspapers and Periodicals. Statistical material on a number of important current socio-economic
problems can be obtained from the numerical data collected and published by some reputed magazines,
periodicals and newspapers like Eastern Economist, Economic Times, The Financial Express, Indian
Journal of Economics, Commerce, Capital, Transport, Statesman’s Year Book and The Times of India Year
Book, etc.
7. International Publications. The publications of a number of foreign governments or international
agencies provide invaluable statistical information on a variety of important economic and current topics.
The publications of the United Nations Organisation (U.N.O.) like U.N.O. Statistical Year Book, U.N.
Statistical Abstract, Demographic Year Book, etc., and its subsidiaries like World Health Organisation
(W.H.O.) on contagious diseases ; annual reports of International Labour Organisation (I.L.O.) ;
International Monetary Fund (I.M.F.) ; World Bank ; Economic and Social Commission for Asia and
Pacific (ESCAP) ; International Finance Corporation (I.F.C.) ; International Statistical Education Institute
and so on are very valued publications of secondary data.
Remark. It may be pointed out that the various publications enumerated above vary as regards the
periodicity of their publications. Some are published periodically at regular intervals of time (such as
weekly, monthly, quarterly or annually) whereas others are ad-hoc publications which do not have any
specific periodicity of publications.
2·5·2. Unpublished Sources. The statistical data need not always be published. There are various
sources of unpublished statistical material such as the records maintained by private firms or business
enterprises who may not like to release their data to any outside agency ; the various departments and
offices of the Central and State Governments ; the researches carried out by the individual research scholars
in the universities or research institutes.
Remark. In some of the socio-economic surveys the information is gathered from the respondents with
the promise that it is exclusively meant for research programmes and will be kept strictly confidential. Such
data are not published. In case it is published, it is done with a brief note namely, “Source : Confidential.”
2·6. PRECAUTIONS IN THE USE OF SECONDARY DATA
Secondary data should be used with extra caution. Before using such data, the investigator must be
satisfied regarding the reliability, accuracy, adequacy and suitability of the data to the given problem under
investigation.
Proper care should be taken to edit it so that it is free from inconsistencies, errors and omissions. In the
words of L.R. Connor “Statistics, especially other peoples’ statistics, are full of pitfalls for the user” and
therefore, secondary data should not be used before subjecting it to a thorough and careful scrutiny.
COLLECTION OF DATA
2·19
Prof. A.L. Bowley also remarks, “It is never safe to take the published statistics at their face value without
knowing their meaning and limitations and it is always necessary to criticise the arguments that can be
based upon them.” In using secondary data we should take a special note of the following factors.
1. The Reliability of Data. In order to know about the reliability of the data, we should satisfy ourselves
about :
(i) the reliability, integrity and experience of the collecting organistion.
(ii) the reliability of source of information and
(iii) the methods used for the collection and analysis of the data.
It should be ascertained that the collecting agency was unbiased in the sense that it had no personal
motives and right from the collection and compilation of the data to the presentation of results in the final
form in the selected source, the data was thoroughly scrutinised and edited so as to make it free from errors
as far as possible. Moreover, it should also be verified that the data relates to normal times free from
periods of economic boom or depression or natural calamities like famines, floods, earthquakes, wars, etc.,
and is still relevant for the purpose in hand.
If the data were collected on the basis of a sample we should satisfy ourselves that :
(i) The sample was adequate (not too small).
(ii) It was representative of the characteristics of the population, i.e., it was selected by proper sampling
technique.
(iii) The data were collected by trained, experienced and unbiased investigators under the proper
supervisory checks on the field work so that sampling errors were minimised.
(iv) Proper estimation techniques were used for estimating the parameters of the population.
(v) The desired degree of accuracy was achieved by the compiler.
Remark. A source note, giving in details the sources from which data were obtained is imperative for
the validity of the secondary data since to the learned users of statistics the reputations of the sources may
vary greatly from one agency to another.
2. The Suitability of Data. Even if the data are reliable in the sense as discussed above it should not be
used without confirming that it is suitable for the purpose of enquiry under investigation. For this, it is
important :
(i) To observe and compare the objectives, nature and scope of the given enquiry with the original
investigation.
(ii) To confirm that the various terms and units were clearly defined and uniform throughout the earlier
investigation and these definitions are suitable for the present enquiry also. For instance, a unit like
household, wages, prices, farm, etc., may be defined in many different ways. If the units are defined
differently in the original investigation than what we want, the secondary data will be termed as unsuitable
for the present enquiry. For example, if we want to construct the cost of living indices, it must be ensured
that the original data relating to prices was obtained from retail shops, co-operative stores or super bazars
and not from the wholesale market.
(iii) To take into account the difference in the timings of collection and homogeneity of conditions for
the original enquiry and the investigation in hand.
3. Adequacy of Data. Even if the secondary data are reliable and suitable in terms of the discussion
above, it may not be adequate enough for the purpose of the given enquiry. This happens when the
coverage given in the original enquiry was too narrow or too wide than what is desired in the current
enquiry or in other words when the original data refers to an area or a period which is much larger or
smaller than the required one. For instance, if the original data relate to the consumption pattern of the
various commodities by the people of a particular State, say, Maharashtra then it will be inadequate if we
want to study the consumption pattern of the people for the whole country. Similarly if the original data
relate to yearly figures of a particular phenomenon, it will be inadequate if we are interested in the monthly
study. This is so because of the fluctuations in the phenomenon in different regions or periods.
Another important factor to decide about the adequacy of the available data for the given investigation
is the time period for which the data are available. For example, if we are given the values of a particular
2·20
BUSINESS STATISTICS
phenomenon (say, profits of a business concern or production of a particular commodity) for the last 3-4
years, it will be inadequate for studying the trend pattern for which the values for the last 8-10 years will be
required.
Hence, in order to arrive at conclusions free from limitations and inaccuracies, the published data i.e.,
the secondary data must be subjected to thorough scrutiny and editing before it is accepted for use.
EXERCISE 2·1.
1. (a) What do you mean by a statistical enquiry ? Describe the main stages in a statistical enquiry.
(b) Describe the process of planning a statistical enquiry, with special reference to its scope and purpose, choice
between sample and census approaches, accuracy and analysis of data.
2. If you are appointed to conduct a statistical enquiry, describe in general, what steps will you be taking from the
stage of appointment till the presentation of your report.
3. (a) Distinguish between (i) primary and secondary data (ii) sampling and census method.
(b) Distinguish between primary data and secondary data and discuss the various methods of collecting primary
data.
[C.S. (Foundation), Dec. 2000]
(c) Explain the different methods of collecting primary data.
(Punjab Univ. B.Com., 1996)
4. (a) What are the various methods of collecting statistical data ? Which of these is most reliable and why ?
(b) Describe the methods generally employed in the collection of statistical data, stating briefly their merits and
demerits.
5. (a) Distinguish between Primary and Secondary data. Give a brief account of the chief methods of collecting
Primary Data and bring out their merits and defects.
(b) Discuss the various methods of collecting ‘primary data’. State the methods you would employ to collect
information about utilisation of plant capacity in small-scale sector in the Union Territory of Delhi.
6. Distinguish between ‘Primary Data’ and ‘Secondary Data’. State the chief sources of Secondary Data. What
precautions are to be observed when such data are to be used for any investigation ?
7. A firm’s own records are internal data. What is meant by external data, a primary data, a primary source and
secondary source ? Which is preferred, primary sources or secondary sources, and why ? Why do you suppose
secondary sources are so often used ?
8. (a) What are consumer primary and secondary data ? State those factors which should be kept in mind while
using secondary data for the investigation.
(b) Distinguish between primary and secondary data. What precautions should be taken in the use of secondary
data ?
(c) “Statistics, especially other people’s statistics are full of pitfalls for the user unless used with caution.”
Elucidate the statement and mention what are the sources of the secondary data.
[Delhi Univ. B.Com. (Pass), 2001]
9. (a) “In collection of statistical data common sense is the chief requisite and experience the chief teacher”.
Discuss the above statement with comments.
(b) “It is never safe to take published statistics at their face value without knowing their meanings and limitations
and it is always necessary to criticise the arguments that can be based on them.” (Bowley). Elucidate.
10. (a) Define a statistical unit and explain what should be the essential requirements of a good statistical unit.
(b) What are the essential points to be remembered in the choice of statistical units ?
(c) What is a statistical unit ? What do you mean by units of collection and units of analysis ? Discuss their relative
uses.
(d) Giving appropriate reasons, state what units can be used for the following :
(i) Production of cotton in textile industry.
(ii) Labour employed in industry.
(iii) Consumption of electricity.
11. What are the essentials of a questionnaire ? Draft a questionnaire not exceeding ten questions to study the
views on educational programmes of television and indicate an outline of the design of the survey.
12. What do you mean by a questionnaire ? What is the difference between a questionnaire and a schedule ? State
the essential points to be remembered in drafting a questionnaire.
13. Discuss the essentials of a good questionnaire. “It is proposed to conduct a sample survey to obtain
information on the study habits of University students in Chandigarh and the facilities available to them”. Explain how
you will plan the survey. Draft a suitable questionnaire for this purpose.
COLLECTION OF DATA
2·21
14. What are the different methods of collection of data ? Why are personal interviews usually preferred to
questionnaire ? Under what conditions may a questionnaire prove as satisfactory as a personal interview ?
15. You are the Sales Promotion Officer of Delta Cosmetics Co. Ltd. Your company is about to market a new
product. Design a suitable questionnaire to conduct a consumer survey before the product is launched. State the various
types of persons that may be approached for replying to the questionnaire.
16. It is required to collect information on the economic conditions of textile mill workers in Mumbai. Suggest a
suitable method for collection of primary data. Draft a suitable questionnaire of about ten questions for collecting this
information. Also suggest how you will proceed to carry out statistical analysis of the information collected.
17. What are the essentials of a good questionnaire ? Draft a suitable questionnaire to enable you to study the
effects of super markets on prices of essential consumer goods.
18. What are the chief features of a good questionnaire ? What precautions do you take while drafting a
questionnaire ?
19. How would you organise an enquiry into the cost of living of the student community in a city ? Draw up a
blank form to obtain the required information.
20. Fill in the blanks :
(i) There are . . . . . . methods of collecting primary data.
(ii) Data are classified into . . . . . . and . . . . . .
(iii) . . . . . . is a suitable method of collecting data in cases where the informants are literate and spread over a
vast area.
(iv) Data originally collected for any investigation is called . . . . . .
(v) The . . . . . . data should be used after careful scrutiny.
Ans. (i) Four ; (ii) Primary and Secondary ; (iii) Mailed questionnaire method; (iv) Primary data; (v) Secondary.
21. What methods would you employ in collection of data considering accuracy, time and cost involved when the
field of enquiry is :
(i) small; (ii) fairly large; (iii) very large.
Ans. (i) Direct personal interview ; (ii) and (iii) Mailed questionnaire method.
22. Assume that you employ the following data while conducting a statistical investigation :
(i) Estimate of personal income taken from R.B.I. bulletin.
(ii) Financial data of Indian companies taken from the annual reports of the Ministry of Law and Company
Affairs.
(iii) Tabulation from schedules used in interviews that you yourself conducted.
(iv) Data collected by the National Sample Survey.
Which of the above is Primary Data ?
Ans. Only (iii).
23. Explain the necessity of editing primary and secondary data and briefly discuss points to be considered while
editing such data.
3
Classification and
Tabulation
3·1. INTRODUCTION — ORGANISATION OF DATA
In the last chapter we described the various methods of collecting data for any enquiry. Unfortunately
the data collected in any statistical investigation, known as raw data, are so voluminous and huge that they
are unwieldy and uncomprehensible. So, having collected and edited the data, the next important step is to
organise it i.e., to present it in a readily comprehensible condensed form which will highlight the important
characteristics of the data, facilitate comparisons and render it suitable for further processing (statistical
analysis) and interpretations.
The presentation of the data is broadly classified into the following two categories:
(i) Tabular Presentation.
(ii) Diagrammatic or Graphic Presentation.
A statistical table is an orderly and logical arrangement of data into rows and columns and it attempts
to present the voluminous and heterogeneous data in a condensed and homogeneous form. But before
tabulating the data, generally, systematic arrangement of the raw data into different homogeneous classes is
necessary to sort out the relevant and significant features (details) from the irrelevant and insignificant
ones.
This process of arranging the data into groups or classes according to resemblances and similarities is
technically called classification. Thus, classification of the data is preliminary to its tabulation. It is thus the
first step in tabulation because the items with similarities must be brought together before the data are
presented in the form of a ‘table’.
On the other hand, the diagrams and graphs are pictorial devices for presenting the statistical data.
However, in this chapter, we shall discuss only ‘Classification and Tabulation’ of the data while
‘Diagrammatic and Graphic Presentation’ of the data will be discussed in next chapter (Chapter 4).
3·2. CLASSIFICATION
It is of interest to give below the following definitions of Classification :
“Classification is the process of arranging data into sequences and groups according to their common
characteristics, or separating them into different but related parts.”—Secrist.
“A classification is a scheme for breaking a category into a set of parts, called classes, according to
some precisely defined differing characteristics possessed by all the elements of the category.”
—Tuttle A.M.
Thus classification impresses upon the ‘arrangement of the data into different classes, which are to be
determined depending upon the nature, objectives and scope of the enquiry. For instances the number of
students registered in Delhi University during the academic year 2002-03 may be classified on the basis of
any of the following criterion :
(i) Sex
(ii) Age
(iii) The state to which they belong
3·2
BUSINESS STATISTICS
(iv) Religion
(v) Different faculties, like Arts, Science, Humanities, Law, Commerce, etc.
(vi) Heights or weights
(vii) Institutions (Colleges) and so on.
Thus the same set of data can be classified into different groups or classes in a number of ways based
on any recognisable physical, social or mental characteristic which exhibits variation among the different
elements of the given data. The facts in one class will differ from those of another class w.r.t. some
characteristic called the basis or criterion of classification.
As an illustration, the data relating to socio-economic enquiry e.g., the family budget data relating to
nature, quality and quantity of the commodities consumed by the group of people together with expenditure
on different items of consumption may be classified under the following major heads :
(i) Food
(ii) Clothing
(iii) Fuel and Lighting
(iv) House rent
(v) Miscellaneous (including items like education, recreation, medical expenses, gifts, newspaper,
washerman, etc.).
Each of the above groups or classes may further be divided into sub-groups or sub-classes. For
example, ‘Food’ may be sub-divided into Cereals (rice, wheat, maize, pulses, etc.); Vegetables; Milk and
milk products ; Oil and ghee ; Fruits and Miscellaneous.
Thus it may be understood that to analyse any statistical data, classification may not be limited to one
criterion or basis only. We might classify the given data w.r.t. two or more criteria or bases simultaneously.
This technique of dividing the given data into different classes w.r.t. more than one basis simultaneously is
called cross-classification and this process of further classification may be carried on as long as there are
possible bases for classification. For instance, the students in the university may be simultaneously
classified w.r.t. sex and faculty or w.r.t. age, sex and religion (three criteria) simultaneously and so on.
3·2·1. Functions of Classification. The functions of classification may be briefly summarised as
follows :
(i) It condenses the data. Classification presents the huge unwieldy raw data in a condensed form
which is readily comprehensible to the mind and attempts to highlight the significant features contained in
the data.
(ii) It facilitates comparisons. Classification enables us to make meaningful comparisons depending on
the basis or criterion of classification. For instance, the classification of the students in the university
according to sex enables us to make a comparative study of the prevalence of university education among
males and females.
(iii) It helps to study the relationships. The classification of the given data w.r.t. two or more criteria,
say, the sex of the students and the faculty they join in the university will enable us to study the relationship
between these two criteria.
(iv) It facilitates the statistical treatment of the data. The arrangement of the voluminous
heterogeneous data into relatively homogeneous groups or classes according to their points of similarities
introduces homogeneity or uniformity amidst diversity and makes it more intelligible, useful and readily
amenable for further processing like tabulation, analysis and interpretation of the data.
3·2·2. Rules for Classification. Although classification is one of the most important techniques for the
statistical treatment and analysis of numerical data, no hard and fast rules can be laid down for it.
Obviously, a technically sound classification of the data in any statistical investigation will primarily
depend on the nature of the data and the objectives of the enquiry. However, consistent with the nature and
objectives of the enquiry, the following general guiding principles may be observed for good classification :
(i) It should be unambiguous. The classes should be rigidly defined so that they should not lead to any
ambiguity. In other words, there should not be any room for doubt or confusion regarding the placement of
the observations in the given classes. For example, if we have to classify a group of individuals as
CLASSIFICATION AND TABULATION
3·3
‘employed’ and ‘un-employed’ : or ‘literate’ and ‘illiterate’ it is imperative to define in clear cut terms as
to what we mean by an employed person and unemployed person ; by a literate person and illiterate person.
(ii) It should be exhaustive and mutually exclusive. The classification must be exhaustive in the sense
that each and every item in the data must belong to one of the classes. A good classification should be free
from the residual class like ‘others’ or ‘miscellaneous’ because such classes do not reveal the
characteristics of the data completely. However, if the classes are very large in number as is the case in
classifying various commodities consumed by people in a certain locality, it becomes necessary to
introduce this ‘residual class’ otherwise the purpose of classification viz., condensation of the data will be
defeated.
Further, the various classes should be mutually disjoint or non-overlapping so that an observed value
belongs to one and only one of the classes. For instance, if we classify the students in a college by sex i.e.,
as males and females, the two classes are mutually exclusive. But if the same group is classified as males,
females and addicts to a particular drug then the classification is faulty because the group “addicts to a
particular drug” includes both males and females. However, in such a case, a proper classification will be
w.r.t. two criteria viz., w.r.t. sex (males and females) and further dividing the students in each of these two
classes into ‘addicts’ and ‘non-addicts’ to the given drug.
(iii) It should be stable. In order to have meaningful comparisons of the results, an ideal classification
must be stable i.e., the same pattern of classification should be adopted throughout the analysis and also for
further enquiries on the same subject. For instance, in the 1961 census, the population was classified w.r.t.
profession in the four classes viz. (i) working as cultivator, (ii) working as agricultural labourer,
(iii) working as household industry, and (iv) others. However, in 1971 census, the classification w.r.t.
profession was as under :
(a) Main Activity : (i) Worker [Cultivator (C), Agricultural labourer (AL), Household industries (HHI),
Other works (OW)].
(b) Broad Category. Non-worker [Household duties (H) ; Student (ST) ; Renteer or Retired person
(R); Dependent, Beggers, Institutions and Others (DBIO)].
Consequently the results obtained in the two censuses cannot be compared meaningfully. Hence,
having decided about the basis of classification in an enquiry, we should stick to it for other related matters
in order to have meaningful comparisons.
(iv) It should be suitable for the purpose. The classification must be in keeping with the objectives of
the enquiry. For instance, if we want to study the relationship between the university education and sex, it
will be futile to classify the students w.r.t. to age and religion.
(v) It should be flexible. A good classification should be flexible in that it should be adjustable to the
new and changed situations and conditions. No classification is good enough to be used for ever ; changes
here and there become necessary with the changes in time and changed circumstances. However, flexibility
should not be interpreted as instability of classification. The classification can be kept flexible by
classifying the given population into some major groups which more or less remain stable and allowing for
adjustment due to changed circumstances or conditions by sub-dividing these major groups into sub-groups
or sub-classes which can be made flexible. Hence, the classification can maintain the character of flexibility
along with stability.
Also see § 3·4·2 (Number of Classes) and § 3·4·3 (Size of Class Intervals).
3·2·3. Bases of Classification. The bases or the criteria w.r.t. which the data are classified primarily
depend on the objectives and the purpose of the enquiry. Generally, the data can be classified on the
following four bases :
(i) Geographical i.e., Area-wise or Regional.
(ii) Chronological i.e., w.r.t. occurrence of time.
(iii) Qualitative i.e., w.r.t. some character or attribute.
(iv) Quantitative i.e., w.r.t. numerical values or magnitudes.
In the following section we shall briefly discuss them one by one.
(i) Geographical Classification. As the name suggests, in this classification the basis of classification
is the geographical or locational differences between the various items in the data like States, Cities,
3·4
BUSINESS STATISTICS
Regions, Zones, Areas, etc. For example, the yield of agricultural output per hectare for different countries
in some given period or the density of the population (per square km) in different cities of India, is given in
the following tables.
TABLE 3·1
AGRICULTURAL OUTPUT
OF DIFFERENT COUNTRIES
(In Kg Per Hectare)
Country
TABLE 3·2
DENSITY OF POPULATION
IN DIFFERENT CITIES OF INDIA
(Per Square Kilometre)
Average Output
India
USA
Pakistan
USSR
China
Syria
Sudan
UAR
City
123·7
581·0
257·5
733·5
269·2
615·2
339·8
747·5
Density of Population
Kolkata
685
Mumbai
654
Delhi
423
Chennai
205
Chandigarh
48
Source : Yojana; Vol. XV, No. 18.
19th Sept. 1971, page 22.
Geographical classifications are usually presented either in an alphabetical order (which is generally
the case in reference tables*) or according to size or values to lay more emphasis on the important area or
region as is done in Table 3·2 (and this is generally the case in summary table*)
TABLE 3·3
(ii) Chronological Classification. Chronological
classification is one in which the data are classified on the
POPULATION OF INDIA (In Crores)
basis of differences in time, e.g., the production of an
Year
Population
industrial concern for different periods ; the profits of a big
1901
23·8
business house over different years ; the population of any
1911
25·0
country for different years. We give in Table 3·3, the
1921
25·2
population of India for different decades.
1931
27·9
1941
31·9
The time series data, which are quite frequent in
1951
36·1
Economic and Business Statistics are generally classified
1961
43·9
chronologically, usually starting with the first period of
1971
54·8
occurrence.
1981
68·3
(iii) Qualitative Classification. When the data are
1991
84·4
classified according to some qualitative phenomena which are
not capable of quantitative measurement like honesty, beauty, employment, intelligence, occupation, sex,
literacy, etc., the classification is termed as qualitative or descriptive or w.r.t. attributes. In qualitative
classification the data are classified according to the presence or absence of the attributes in the given units.
If the data are classified into only two classes w.r.t. an attribute like its presence or absence among the
various units, the classification is termed as simple or dichotomous. Examples of such classification are
classifying a given population of individuals as honest or dishonest ; male or female ; employed or
unemployed ; beautiful or not beautiful and so on. However, if the given population is classified into more
than two classes w.r.t. a given attribute, it is said to be manifold classification. For example, for the
attribute intelligence the various classes may be, say, genius, very intelligent, average intelligent, below
average and dull as given below :
Population
↓
↓
↓
Highly Intelligent
Average Intelligent
———————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————
↓
Genius
↓
Below Average
↓
Dull
————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————
* For detailed discussion on reference tables and summary tables, see § 3.7.3, on Types of Tabulation in this
chapter.
3·5
CLASSIFICATION AND TABULATION
Moreover, if the given population is divided into classes on the basis of simultaneous study of more
than one attribute at a time, the classification is again termed as manifold classification. As an illustration,
suppose we classify the population by sex into two classes, males and females and each of these two classes
is further divided into two classes w.r.t. another attribute, say, smoking i.e., smokers and non-smokers, thus
giving us four classes in all. Each of these four classes may further be divided w.r.t. a third attribute, say,
religion into two classes Hindu, Non-Hindu and so on. The scheme is explained below.
Population
↓
—————————————————————————————————————————————————————————————————————————
↓
↓
Male
Female
———————————————————————————————————————
————————————————————————————————————————
↓
↓
↓
↓
Smoker
Non-Smoker
Smoker
Non-Smoker
↓
——————————————————
↓
Hindu
↓
——————————————————
↓
↓
Non-Hindu
Hindu
↓
Non-Hindu
↓
—————————————————————
↓
Hindu
↓
—————————————————————————————
↓
↓
Non-Hindu
Hindu
↓
Non-Hindu
(iv) Quantitative Classification. If the data are
TABLE 3·4
classified on the basis of phenomenon which is capable of DAILY EARNINGS (IN ’00 RUPEES)
quantitative measurement like age, height, weight, prices,
OF 60 DEPARTMENTAL STORES
production, income, expenditure, sales, profits, etc., it is
Number of stores
Daily earnings
termed as quantitative classification. The quantitative
Up to 100
6
phenomenon under study is known as variable and hence this
101—200
14
classification is also sometimes called classification by
201—300
8
variables. For example, the earnings of different stores may be
301—400
10
classified as given in Table 3·4.
401—500
8
In the above classification, the daily earnings of the stores
501—600
6
601—700
4
are termed as variable and the number of stores in each class
701—800
4
as the frequency. The above classification is termed as
grouped frequency distribution.
Variable. As already pointed out, the quantitative phenomenon under study, like marks in a test,
heights or weights of the students in a class, wages of workers in a factory, sales in a departmental store,
etc., is termed a variable or a variate. It may be noted that different variables are measured in different
units e.g., age is measured in years, height in inches or cms ; weight in lbs or kgs, income in rupees and so
on.
Variables are of two kinds :
(i) Continuous variable.
(ii) Discrete variable (Discontinuous variable).
Those variables which can take all the possible values (integral as well as fractional) in a given
specified range are termed as continuous variables. For example, the age of students in a school (Nursery to
Higher Secondary) is a continuous variable because age can take all possible values (as it can be measured
to the nearest fraction of time : years, months, days, minutes, seconds, etc.), in a certain range, say, from
3 years to 20 years. Some other examples of continuous variable are height (in cms), weight (in lbs),
distance (in kms). More precisely a variable is said to be continuous if it is capable of passing from any
given value to the next value by infinitely small gradations.
On the other hand those variables which cannot take all the possible values within a given specified
range are termed as discrete (discontinuous) variables. For example, the marks in a test (out of 100) of a
group of students is a discrete variable since in this case marks can take only integral values from 0 to 100
(or it may take halves or quarters also if such fractional marks are given. Usually, fractional marks, if any,
are rounded to the nearest integer). It cannot take all the values (integral as well as fractional) from 0 to
100. Some other examples of discrete variable are family size (members in a family), the population of a
city, the number of accidents on the road, the number of typing mistakes per page and so on. A discrete
variable is, thus, characterised by jumps and gaps between the one value and the next. Usually, it takes
3·6
BUSINESS STATISTICS
integral values in a given range which depends on the variable under study. A detailed discussion of the
quantitative classification of a series or set of observations is given in § 3·3, “Frequency Distributions”.
Remark. Values of the variable in a ‘given specified range’ are determined by the nature of the
phenomenon under study. In case of heights of students in a college the range may be 4′ 6′′ (i.e., 137 cms)
to 6′ 3′′ (i.e., 190 cms) ; in case of weights, it may be from 100 lbs to 200 lbs, say ; in case of marks in a
test out of 25, it will be from 0 to 25 and so on.
3·3. FREQUENCY DISTRIBUTION
The organisation of the data pertaining to a quantitative phenomenon involves the following four
stages :
(i) The set or series of individual observations - unorganised (raw) or organised (arrayed) data.
(ii) Discrete or ungrouped frequency distribution.
(iii) Grouped frequency distribution.
(iv) Continuous frequency distribution.
We shall explain the various stages by means of a numerical illustration.
Let us consider the following distribution of marks of 200 students in an examination, arranged serially
in order of their roll numbers.
70
55
51
42
57
45
60
47
63
53
33
65
39
78
55
64
58
61
65
42
TABLE 3·5. MARKS OF
50
25
65
75
52
36
45
42
53
59
49
41
45
63
54
52
45
39
64
35
200 STUDENTS
30
20
41
35
40
30
15
53
43
48
46
46
26
18
41
49
37
22
42
48
42
38
41
25
45
36
40
50
52
30
41
32
17
38
28
29
49
46
46
31
46
48
32
40
17
40
50
31
50
42
32
43
42
27
57
34
55
34
47
35
44
43
34
34
38
42
61
43
33
21
33
42
34
51
17
16
42
37
71
37
22
42
39
67
31
37
44
62
38
42
39
42
39
51
49
16
53
52
17
41
53
33
32
50
38
48
37
24
26
40
21
35
38
15
37
28
29
38
23
40
54
39
32
44
17
35
41
33
34
33
39
48
24
33
46
31
53
43
47
36
48
34
39
42
23
38
27
53
53
19
59
57
54
37
47
51
27
59
39
29
The data in the above form is called the raw or disorganised data. In the raw form the data are so
unwieldy and scattered that even after a very careful perusal, the various details contained in them remain
unfollowed and uncomprehensive. The above presentation of the data in its raw form does not give us any
useful information and is rather confusing to the mind. Our objective will be to express the huge mass of
data in a suitable condensed form which will highlight the significant facts and comparisons and furnish
more useful information without sacrificing any information of interest about the important characteristics
of the distribution.
3·3·1. Array. A better presentation of the above raw data would be to arrange them in an ascending or
descending order of magnitude which is called the ‘arraying’ of the data. However, this presentation
(arraying), though better than the raw data does not reduce the volume of the data.
3·3·2. Discrete or Ungrouped Frequency Distribution. A much better way of the representation of
the data is to express it in the form of a discrete or ungrouped frequency distribution where we count the
number of times each value of the variable (marks in the above illustration) occurs in the above data. This
is facilitated through the technique of Tally Marks or Tally Bars as explained below :
In the first column we place all the possible values of the variable (marks in the above case). In the
second column a vertical bar (|) called the Tally Mark is put against the number (value of the variable)
whenever it occurs. After a particular value has occurred four times, for the fifth occurrence we put a cross
3·7
CLASSIFICATION AND TABULATION
tally mark ( ) on the first four tally marks like |||| to give us a block of 5. When it occurs for the 6th time
we put another tally mark against it (after leaving some space from the first block of 5) and for the 10th
occurrence we again put a cross tally mark ( ) on the 6th to 9th tally marks to get another block of 5 and
so on. This technique of putting cross tally marks at every 5th repetition (giving groups of 5 each)
facilitates the counting of the number of occurrences of the value at the end. In the absence of such cross
tally marks we shall get continuous tally bars like ||||||||…and there may be confusion in counting and we are
liable to commit mistakes also. Thus, the 2nd column consists of tally marks or tally bars. After putting
tally marks for all the values in the data, we count the number of times each value is repeated and write it
against the corresponding number (value of the variable) in the third column, entitled frequency. This type
of representation of the data is called discrete or ungrouped frequency distribution. The marks (which vary
from student to student) are called the variable under study and the number of students against the
corresponding marks (which tell us how frequently the marks occur) is called the frequency (f) of the
variable. The Table 3·6 below gives the ungrouped frequency distribution of the data in Table 3·5, along
with the tally marks.
TABLE 3·6. FREQUENCY DISTRIBUTION OF MARKS OF 200 STUDENTS
Marks
15
Frequency
2
Marks
33
|||| ||
Frequency
7
Marks
51
||
2
34
35
18
19
||||
|
|
5
|||| ||
7
52
||||
4
53
3
7
54
55
|||| |||
|||
|||
8
36
37
||||
|||
|||| ||
5
1
1
3
3
20
|
1
38
|||| |||
8
57
|||
3
21
22
||
2
39
|||| ||||
9
58
|
1
||
2
40
|||| |
6
59
|||
3
23
||
2
41
|||| ||
24
||
2
42
|||| |||| ||||
25
||
2
43
26
27
||
|||
2
3
44
45
||||
|||
||||
28
||
2
46
29
30
|||
|||
3
3
47
48
31
32
||||
||||
4
5
49
50
16
17
Tally Bars
||
Tally Bars
|||| |
||||
|||| |
||||
||||
Tally Bars
||||
Frequency
4
7
60
|
1
14
61
||
2
5
62
|
1
3
5
63
64
||
||
2
2
6
65
|||
3
4
6
67
70
|
||
1
2
4
5
75
78
|
|
1
1
From the frequency table given above, we observe that there are 8 students getting 38 marks, 14
students getting 42 marks, only 1 student getting 75 marks and so on. The presentation of the data in the
form of an ungrouped frequency distribution as given above is better way than ‘arraying’ but still it does
not condense the data much and is quite cumbersome to grasp and comprehend. The ungrouped frequency
distribution is quite handy (i) if the values of the variable are largely repeated otherwise there will be hardly
any condensation or (ii) if the variable (X) under consideration takes only a few values, say, if X were the
marks out of 10 in a test given to 200 students, then X is a variable taking the values in the range 0 to 10
and can be conveniently represented by an ungrouped frequency distribution. However, if the variable takes
the values in a wide (large) range as in the above illustration in Table 3·6, the data still remain unwieldy
and need further processing for statistical analysis.
3·3·3. Grouped Frequency Distribution. If the identity of the units (students in our example) about
whom a particular information is collected (marks in the above illustration) is not relevant, nor is the order
in which the observations occur, then the first real step of condensation consists in classifying the data into
different classes (or class intervals) by dividing the entire range of the values of the variable into a suitable
3·8
BUSINESS STATISTICS
number of groups called classes and then recording the number of observations in each group (or class).
Thus, in the above data of Table 3·6, if we divide the total range of the values of the variable viz., 78 – 15 =
63 into groups of size 5 each, then we shall get (63/5) = 13 groups and the distribution of marks is then
given by the following grouped frequency distribution.
TABLE 3·7. DISTRIBUTION OF MARKS OF 200 STUDENTS
Marks (X)
15—19
20—24
25—29
30—34
35—39
40—44
45—49
No. of Students (f)
11
9
12
26
32
35
25
Marks (X)
50—54
55—59
60—64
65—69
70—74
75—79
No. of Students (f)
24
10
8
4
2
2
The various groups into which the values of the variable are classified are known as classes or class
intervals ; the length of the class interval (which is 5 in the above case) is called the width or magnitude of
the classes. The two values specifying the class are called the class limits ; the larger value is called the
upper class limit and the smaller value is called the lower class limit.
3·3·4. Continuous Frequency Distribution. While dealing with a continuous variable it is not
desirable to present the data into a grouped frequency distribution of the type given in Table 3·7. For
example, if we consider the ages of a group of students in a school, then the grouped frequency distribution
into the classes 4—6, 7—9, 10—12, 13—15, etc., will not be correct, because this classification does not
take into consideration the students with ages between 6 and 7 years i.e., 6 < X < 7 ; between 9 and 10 years
i.e., 9 < X < 10 and so on. In such situations we form continuous class intervals, (without any gaps), of the
following type :
Age in years :
Below 6
6 or more but less than 9
9 or more but less than 12
12 or more but less than 15
and so on, which takes care of all the students with any fractions of age.
The presentation of the data into continuous classes of the above type along with the corresponding
frequencies is known as continuous frequency distribution. [For further detailed discussion, see Types of
Classes—Inclusive and Exclusive—§ 3·4·4, page 3.11].
3·4. BASIC PRINCIPLES FOR FORMING A GROUPED FREQUENCY DISTRIBUTION
In spite of the great importance of classification in statistical analysis, no hard and fast rules can be laid
down for it. A statistician uses his discretion for classifying a frequency distribution, and sound experience,
wisdom, skill and aptness are required for an appropriate classification of the data. However, the following
general guidelines may be borne in mind for a good classification of the frequency data.
3·4·1. Types of Classes. The classes should be clearly defined and should not lead to any ambiguity.
Further, they should be exhaustive and mutually exclusive (i.e., non-overlapping) so that any value of the
variable corresponds to one and only one of the classes. In other words, there is one to one correspondence
between the value of the variable and the class.
3·4·2. Number of Classes. Although no hard and fast rule exists, a choice about the number of classes
(class intervals) into which a given frequency distribution can be divided primarily depends upon :
(i) The total frequency (i.e., total number of observations in the distribution),
(ii) The nature of the data i.e., the size or magnitude of the values of the variable,
(iii) The accuracy aimed at, and
3·9
CLASSIFICATION AND TABULATION
(iv) The ease of computation of the various descriptive measures of the frequency distribution such as
mean, variance, etc., for further processing of the data.
However, from practical point of view the number of classes should neither be too small nor too large.
If too few classes are used, the classification becomes very broad and rough in the sense that too many
frequencies will be concentrated or crowded in a single class. This might obscure some important features
and characteristics of the data, thereby resulting in loss of information. Moreover, with too few classes the
basic assumption that class marks (i.e., mid-values of the classes) are representative of the class for
computation of further descriptive measures of distribution like mean, variance, etc., will not be valid, and
the so-called grouping error will be larger in such cases. Consequently, in general, the accuracy of the
results decreases as the number of classes becomes smaller and smaller. On the other hand, too many
classes i.e., large number of classes will result in too few frequencies in each class. This might give
irregular pattern of frequencies in different classes thus making the frequency distribution (frequency
polygon) irregular. Moreover a large number of classes will render the distribution too unwieldy to handle,
thus defeating the very purpose (or aim viz., summarisation of the data) of classification. Further the
computational work for further processing of the data will unnecessarily become quite tedious and time
consuming without any proportionate gain in the accuracy of the results. However, a balance should be
struck between these two factors viz., the loss of information in the first case (i.e., too few classes) and
irregularity of frequency distribution in the second case (i.e., too many classes) to arrive at a pleasing
compromise, giving the optimum number of classes in the view of the statistician. Ordinarily, the number
of classes should not be greater than 20 and should not be less than 5, of course keeping in view the points
(i) to (iv) given above together with the magnitude of class interval, since the number of classes is inversely
proportional to the magnitude of the class interval.
A number of rules of the thumb have been proposed for calculating the proper number of classes.
However, an elegant, though approximate formula seems to be one given by Prof. Sturges known as
Sturges’ rule, according to which
k = 1 + 3·322 log10 N
…(3·1)
where k is the number of class intervals (classes) and N is the total frequency i.e., total number of
observations in the data. The value obtained in (3·1) is rounded to the next higher integer.
Since log of one digited number is 0· (…) ; log of two digited number is 1· (…) ; log of three digited
number is 2· (…) and log of four digited number is 3· (…), the use of formula (3·1) restricts the value of k,
the number of classes, to be fairly reasonable. For example :
If N
If N
If N
If N
If N
= 10,
= 100,
= 500,
= 1000,
= 10000,
_4
k = 1 + 3·322 log10 10 = 4·322 ~
[·.· loga a = 1]
k = 1 + 3·322 log10 100 = 1 + 3·322 × 2 log10 10 = 1 + 6·644 = 7·644 _~ 8
_ 10
k = 1 + 3·322 log10 500 = 1 + 3·322 × 2·6990 = 1 + 8·966 = 9·966 ~
k = 1 + 3·322 log10 1000 = 1 + 3·322 × 3 log10 10 = 1 + 9·966 = 10·966 _~ 11
k = 1 + 3·322 × 4 = 1 + 13·288 = 14·288 _~ 14
Accordingly, the Sturges’ formula (3·1) very ingeniously restricts the number of classes between 4 and
20, which is a fairly reasonable number from practical point of view.
The rule, however, fails if the number of observations is very large or very small.
Remarks. 1. The number of class intervals should be such that they usually give uniform and unimodal
distribution in the sense that the frequencies in the given classes first increase steadily, reach a maximum
and then decrease steadily. There should not be any sudden jumps or falls which result in the so-called
irregular distribution. The maximum frequency should not occur in the very beginning or at the end of the
distribution nor should it (maximum frequency) be repeated in which cases we shall get an irregular
distribution.
2. The number of classes should be a whole number (integer) preferably 5 or some multiple of 5 viz.,
10, 15, 20, 25, etc., which are readily perceptible to the mind and are quite convenient for numerical
computations in the further processing (statistical analysis) of the data. Uncommon figures like 3, 7, 11,
etc., should be avoided as far as possible.
3·10
BUSINESS STATISTICS
3·4·3. Size of Class Intervals. Since the size of the class interval is inversely proportional to the
number of classes (class intervals) in a given distribution, from the above discussion it is obvious that a
choice about the size of the class interval will also largely depend on the sound subjective judgement of the
statistician keeping in mind other considerations like N (total frequency), nature of the data, accuracy of the
results and computational ease for further processing of the data. Here an approximate value of the
magnitude (or width) of the class interval, say, ‘i’ can be obtained by using Sturges’ rule (3.1) which gives :
i=
Range
Range
=
Number of Classes 1 + 3·322 log N
10
[Using (3·1)]
where Range of the distribution is given by the difference between the largest (L) and the smallest (S) value
in the distribution.
i.e.,
Range = Xmax – Xmin = L – S
…(3·2)
Range
L–S
∴
i =
=
…(3·3)
N
k
1 + 3·322 log
10
Another ‘rule of the thumb’ for determining the size of the class interval is that : “The length of the
class interval should not be greater than 14 th of the estimated population standard deviation.”* Thus, if ^σ is
the estimate of the population standard deviation then the length of class interval is given by
^ /4 = A, (say).
i ≤ σ
Remarks 1. From (3·3), we get :
k =
Range
Range
≥
i
A
…(3·4)
[
·.· i ≤ A ⇒
1 1
≥
i
A
]
… (3·5)
Thus, (3·4) also enables us to have an idea about the minimum number of classes (k) which will be
given by :
Range
k =
…(3·5)
A
^ /4.
where range is defined in (3·2) and A = σ
If we consider a hypothetical frequency distribution of the life time of 400 radio bulbs tested at a
certain company with the result that minimum life time is 340 hours and maximum life time is 1300 hours
such that, in usual notations :
N = 400, L = 1300 hrs, S = 340 hrs, then using (3·3), we get
1300 – 340
i = 1 + 3·322
log 400
10
960
960
960
= 1 + 3·322 × 2·6021 = 1 + 8·644 = 9·644
= 99·54 –~ 100
…(*)
If the magnitude of the class interval is taken as 100, then the number of classes will be 10 [which is
nothing but the value 9·644 –~ 10 in the denominator of (*)].
2. Like the number of classes, as far as possible, the size of class intervals should also be taken as 5 or
some multiple of 5 viz., 10, 15, 20, etc., for facilitating computations of the various descriptive measures of
the frequency distribution like mean (x–), standard deviation (σ), moments, etc.
3. Class intervals should be so fixed that each class has a convenient mid-point about which all the
observations in the class cluster or concentrate. In other words, this amounts to saying that the entire
frequency of the class is concentrated at the mid-value of the class. This assumption will be true only if the
frequencies of the different classes are uniformly distributed in the respective class intervals. This is a very
fundamental assumption in the statistical theory for the computation of various statistical measures, like
mean, standard deviation, etc.
4. From the point of view of practical convenience, as far as possible, it is desirable to take the class
intervals of equal or uniform magnitude throughout the frequency distribution. This will facilitate the
————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————
* For detailed discussion on “Standard Deviation” see Chapter 6.
3·11
CLASSIFICATION AND TABULATION
computations of various statistical measures and also result in meaningful comparisons between different
classes and different frequency distributions. Further, frequency distributions with equal classes can be
represented diagrammatically with greater ease and utility whereas in the case of classes with unequal
widths the diagrammatic representation might give a distorted picture and thus lead to fallacious
interpretations. However, it may not be practicable nor desirable to keep the magnitudes of the class
intervals equal if there are very wide gaps in the observed data e.g., in the frequency distribution of
incomes, wages, profits, savings, etc., For example, in the frequency distribution of income, larger class
intervals would (obscure) sacrifice all the details about the smaller incomes and smaller classes would give
quite an unwieldy frequency distribution. Such distributions are quite common in many economic and
medical data, where we have to be content with classes of unequal width.
3·4·4. Types of Class Intervals. As already stated, each class is specified by two extreme values called
the class limits, the smaller one being termed as the lower limit and the larger one the upper limit of the
class. The classification of a frequency distribution into various classes is of the following types :
(a) Inclusive Type Classes. The classes of the type 30—39, 40—49, 50—59, 60—69, etc., in which
both the upper and lower limits are included in the class are called “inclusive classes”. For instance, the
class interval 40—49 includes all the values from 40 to 49, both inclusive. The next value viz., 50 is
included in the next class 50—59 and so on. However, the fractional values between 49 and 50 cannot be
accounted for in such a classification. Hence, ‘Inclusive Type’ of classification may be used for a grouped
frequency distribution for discrete variables like marks in a test, number of accidents on the road, etc.,
where the variable takes only integral values. It cannot be used with advantage for the frequency
distribution of continuous variables like age, height, weight, etc., where all values (integral as well as
fractional) are permissible.
(b) Exclusive Type Classes. Let us consider the distribution of ages of a group of persons into classes
15—19, 20—24, 25—29, etc., each of magnitude 5. This classification of ‘inclusive type’ for ages is
defective in the sense that it does not account for the individuals with ages more than 19 years but less than
20 years. In such a situation (where the variable is continuous), the classes have to be made without any
gaps as given below :
15 years and over but under 20
…(*)
20 years and over but under 25
25 years and over but under 30
and so on ; each class in this case also being of magnitude 5. More precisely the above classes can be
written as :
15—20
i.e.,
15 ≤ X < 20
20—25
i.e.,
20 ≤ X < 25
…(**)
25 ≤ X < 30
25—30
i.e.,
}
}
and so on, where it should be clearly understood that in the above classes, the upper limits of each class are
excluded from the respective classes. Such classes in which upper limits are excluded from the respective
classes and are included in the immediate next class are termed as ‘exclusive classes’.
Remarks 1. For ‘exclusive classes’ the presentation given in (*) is preferred since it does not lead to
any confusion. However, if presentation (**) is used, there is slight confusion about the overlapping values
viz., 20, 25, 30, etc., but whenever such presentation is used (which is extensively done in practice) it
should be clearly understood that the upper limit of the class
TABLE 3·8
is to be excluded from that class. From the above discussion
Age (on last birthday) No. of persons (f)
it is also clear that a choice between the ‘inclusive method’
20—24
6
or ‘exclusive method’ of classification will depend on the
25—29
10
nature of the variable under study. For a discrete variable,
30—34
14
the ‘inclusive classes’ may be used while for continuous
35—39
9
variable the ‘exclusive classes’ are to be used.
40—44
6
2. However, sometimes, even for a continuous random
45—49
5
variable the classification may be given to be of ‘inclusive
type’. As an illustration, let us consider the frequency distribution of age of a group of 50 individuals given
in Table 3·8.
3·12
BUSINESS STATISTICS
Although the variable (age) X is a continuous variable, here inclusive type of classes are used since we
are recording the age as on late birthday and consequently it becomes a discrete variable taking only
integral values. Since age is a continuous variable, we might like to convert this ‘inclusive type’
classification into ‘exclusive type’ classification. Since the ages are recorded as on last birthday, they are
recorded almost one year younger (prior). For example, in the
TABLE 3·9
age group 20—24, there may be a person (or many persons) with
Age (on last birthday)
No. of persons
ages 24·1, 24·2,...........up to 24·99. Thus all these persons who
(X)
(f)
have not yet completed 25 years will be taken in the age group
20—25
6
20—24. Hence for obtaining ‘exclusive classes’ we can make
25—30
10
a correction in the above distribution by converting 24 to 25.
30—35
14
Accordingly for continuous representation of data (exclusive
35—40
9
type), all the upper class limits in Table 3·8 will have to be
40—45
6
increased by 1, thereby giving the (exclusive type) distribution,
45—50
5
as given in Table 3·9.
TABLE 3·10
However, if the variable X is taken to denote the ‘age on
Age
(on
next
birthday)
No. of persons
next birthday’, then it would imply that the ages are recorded
(X)
(f)
one year advance (i.e., one year older than existing one). This
will mean that the class 20—25 may include person(s) with
19—24
6
24—29
10
ages just higher than 19 also. As such for continuity of the data
29—34
14
the lower limit will have to be reduced by 1. Hence to obtain
34—39
9
the ‘exclusive type’ classification for this case (X -age on next
39—44
6
birthday i.e., coming birthday), we shall have to subtract 1 from
44—49
5
the lower limit of each class in Table 3·8 to get the distribution
as given in Table 3·10.
3. As far as possible the class limits should start with zero or some convenient multiple of 5. As an
illustration if we want to form a frequency distribution of wages in a factory with class interval of 10 and
the lowest value of wages (per week) is given to be Rs. 43, then instead of having classes 43—53,
53—63,…etc., a proper classification should be 40—50, 50—60, etc.
4. Class Boundaries. If in a grouped frequency distribution there are gaps between the upper limit of
any class and lower limit of the succeeding class (as in the case of inclusive type of classification), there is
need to convert the data into a continuous distribution by applying a correction for continuity for
determining new classes of exclusive type. The upper and lower class limits of the new ‘exclusive type’
classes as called class boundaries.
If d is the gap between the upper limit of any class and lower limit of the succeeding class, the class
boundaries for any class are then given by :
Upper class boundary = Upper class limit + 12 d
…(3·6)
1
Lower class boundary = Lower class limit – 2 d
}
d/2 is called the correction factor.
As an illustration, consider the following distribution of marks :
TABLE 3·11
Marks
20—24
25—29
30—34
35—39
40—44
20
25
30
35
40
–
–
–
–
–
0·5,
0·5,
0·5,
0·5,
0·5,
Class Boundary
24 + 0·5 i.e.,
29 + 0·5 i.e.,
34 + 0·5 i.e.,
39 + 0·5 i.e.,
44 + 0·5 i.e.,
19·5,
24·5,
29·5,
34·5,
39·5,
24·5
29·5
34·5
39·5
44·5
d = 0·5
2
This technique enables us to convert a grouped frequency distribution (inclusive type) into continuous
frequency distribution and is extensively helpful in computing certain statistical measures like mode,
median, etc., [See Chapter 5] which require the distribution to be continuous.
Here, d = 25 – 24 = 30 – 29 = 35 – 34 = 1
⇒
3·13
CLASSIFICATION AND TABULATION
Thus, in Table 3·11, the lower class limits are 20, 25, 30, …, 40 and the upper class limits are 24,
29,…, 44, while the lower class boundaries are 19·5, 24·5,…, 39·5 and the upper class boundaries are 24·5,
29·5,…, 44·5.
5. Mid-value or Class Mark. As the name suggests, the mid-value or the class-mark is the value of
the variable which is exactly at the middle of the class. The mid-value of any class is obtained on dividing
the sum of the upper and lower class limits (or class boundaries) by 2. In other words :
Mid-value of a class = 12 [Lower class limit + Upper class limit]
…(3·7)
= 21 [Lower class boundary + Upper class boundary]
}
In the Table 3.11 of remark 4, it may be seen that the mid-values of various classes are : 22, 27, 32, 37,
42 respectively as given below :
1
2
1
2
1
2
1
2
1
2
}
1
}
(20 + 24) = 22
2 (19·5 + 24·5) = 22
1
(25 + 29) = 27
2 (24·5 + 29·5) = 27
1
(30 + 34) = 32
or
2 (29·5 + 34·5) = 32
1
(35 + 39) = 37
2 (34·5 + 39·5) = 37
1
(40 + 44) = 42
2 (39·5 + 44·5) = 42
It may be noted that whether we use class limits or class boundaries, the mid-values remain same.
Important Note. For fixing the class limits the most important factor to be kept in mind is as given
below:
“The class limits should be chosen in such a manner that the observations in any class are evenly
distributed throughout the class interval so that the actual average of the observations in any class is very
close to the mid-value of the class. In other words, this amounts to saying that the observations are
concentrated at the mid points of the classes.”
This is a very fundamental assumption in preparing a grouped or continuous frequency distribution for
computation of various statistical measures like mean, variance, moments, etc. [See Chapters 5, 6, 7] for
further analysis of the data. If this assumption is not true then the classification will not reveal the main
characteristics and thus give a distorted picture of the distribution. The deviation from this assumption
introduces the so-called ‘grouping error’.
(c) Open End Classes. The classification is termed as ‘open end classification’ if the lower limit of the
first class or the upper limit of the last class are not specified and such classes in which one of the limits is
missing are called ‘open end classes’. For example, the classes like the marks less than 20; age above 60
years, salary not exceeding Rupees 100 or salaries over Rupees 200, etc., are ‘open end classes’ since one
of the class limits (lower or upper) is not specified in them. As far as possible, open end classes should be
avoided since in such classes the mid-value or class-mark cannot be accurately obtained and this poses
problems in the computation of various statistical measures for further processing of the data. Moreover,
open end classes present problems in graphic presentation of the data also.
However, the use of open end classes is inevitable or unavoidable in a number of practical situations,
particularly relating to economic and medical data where there are a few observations with extremely small
or large values while most of the other observations are more or less concentrated in a narrower range.
Thus, we have to resort to open end classes for the frequency distribution of income, wages, profits,
payment of income-tax, savings, etc.
Remark. In case of open end classes, it is customary to estimate the class-mark or mid-value for the
first class with reference to the succeeding class (i.e., 2nd class). In other words, we assume that the
magnitude of the first class is same as that of second class. Similarly, the mid-value of the last class is
determined with reference to the preceding class i.e., last but one class. This assumption will, of course,
introduce some error in the calculation of further statistical measures (averages, dispersion, etc.—See
Chapters 5, 6). However, if only a few items fall in the open end classes then :
(i) there won’t be much loss in information in further processing of data as a consequence of open end
classes, and
(ii) the open end classes will not seriously reduce the utility of graphic presentation of the data.
3·14
BUSINESS STATISTICS
Example 3·1. Form a frequency distribution from the following data by Inclusive Method, taking 4 as
the magnitude of class-intervals :
17,
10,
15,
22,
11,
16,
19,
24,
29,
18,
25,
26,
32,
14,
20,
17,
23,
27,
30,
12,
15,
18,
24,
36,
18,
15,
21,
28,
38,
33,
34,
13,
10,
16,
20,
22,
29,
19,
23,
31.
TABLE 3·12
Solution. Since the minimum value of the variable is 10
FREQUENCY DISTRIBUTION
which is a very convenient figure for taking the lower limit of
the first class and the magnitude of the class intervals is given Class Interval Tally Marks Frequency (f)
to be 4, the classes for preparing frequency distribution by
10—13
||||
5
the ‘Inclusive Method’ will be 10—13, 14—17, 18—21,
14—17
8
||||
|||
22—25,…, 34—37, 38—41, the last class being 38—41,
18—21
8
||||
|||
because the maximum value in the distribution is 38.
22—25
7
|||| ||
To prepare the frequency distribution, since the first
26—29
5
||||
value 10 occurs in class 10—13 we put a tally mark against it,
30—33
4
||||
for the value 17 we put a tally mark against the class 14—17 ;
34—37
2
||
for the value 15 we put a tally mark against the class 14—17
38—41
1
|
and so on. The final frequency distribution along with the
tally marks is given in Table 3·12.
Total 40
Example 3·2. Following figures relate to the weekly wages of workers in a factory.
100
75
79
80
110
93
109
84
95
77
100
89
84
81
106
96
94
83
95
78
101
99
83
89
102
97
93
82
97
80
102
96
87
99
107
99
97
80
98
93
Wages (in ’00 Rs.)
106
86
94
93
88
89
104
100
103
101
100
102
98
99
84
86
100
105
96
97
82
92
75
103
101
103
100
88
106
98
87
90
76
104
101
107
97
91
103
98
109
86
76
107
86
107
88
93
85
98
104
78
79
110
94
108
86
95
84
87
Prepare a frequency table by taking a class interval of 5.
Solution. In the above distribution, the minimum value of the variable X (wages in ’00 Rupees) is 75
and the maximum value is 110. Moreover, the magnitude of the class intervals is given to be 5. Since
‘wages’ is a continuous variable, the frequency distribution with ‘Exclusive Method’ would be appropriate.
Since the minimum value 75 is a convenient figure to be taken as the lower limit of the first class, the class
intervals may be taken as 75—80, 80—85, 85—90,…, 110—115, the upper limit of each class being
included in the next class. The frequency distribution is given in Table 3·13.
TABLE 3·13
FREQUENCY DISTRIBUTION OF WAGES OF WORKERS IN A FACTORY
Weekly Wages (in ’00 Rs.) (X)
75—80
80—85
85—90
90—95
95—100
100—105
105—110
110—115
Tally Marks
||||
||||
||||
||||
||||
||||
||||
||
||||
||||
||||
||||
||||
||||
||||
||
||||
|
||||
||||
|
No. of Workers (f)
||||
||||
9
12
15
11
20
20
11
2
Total = 100
3·15
CLASSIFICATION AND TABULATION
Example 3·3. Prepare a frequency distribution of the number of letters in a word from the following
excerpt (ignore punctuation marks).
“In the beginning”, said a Persian Poet, “Allah took a rose, a lily, a dove, a serpent, a little honey, a
Dead Sea Apple and a handful of clay. When he looked at the amalgam – it was a woman.”
Also obtain (i) the number of words with 6 letters or more, (ii) the proportion of words with 5 letters or
less, and (iii) the percentage of words with number of letters between 2 and 8 (i.e., more than 2 but less
than 8).
Solution. Let X denote the number of letters in each word in the excerpt given above. We note that in
the above excerpt there are words with number of letters ranging from 1 to 9. Hence X takes the values
from 1 to 9. For example, in the first word ‘In’ there are 2
TABLE 3·14
letters ; in the second world ‘the’ there are 3 letters ; in the
FREQUENCY DISTRIBUTION
third word ‘beginning’ there are 9 letters and so on. Thus,
the corresponding values of the variable X in the above OF NUMBER OF LETTERS IN A WORD
excerpt are as given below :
Number of letters Tally Marks Frequency (f)
in a word (X)
2, 3, 9, 4, 1, 7, 4,
5, 4, 1, 4, 1, 4
1, 4, 1, 7, 1, 6, 5,
1, 4, 3, 5, 3, 1
7, 2, 4, 4, 2, 6, 2,
3, 7, 2, 3, 1, 5
The frequency distribution along with the tally marks is
given in the Table 3·14.
(i) The number of words with 6 letters or more
=2+4+1=7
(ii) The proportion of words with 5 letters or less is
given by :
39 – 7 32
39 = 39 =
(
5 + 9 + 4 32
0·82 or 9 + 5 +39
= 39 = 0·82
1
2
3
4
5
6
7
8
9
9
5
5
9
4
2
4
0
1
|||| ||||
||||
||||
|||| ||||
||||
||
||||
—
|
Total 39
)
(iii) The percentage of words with the number of letters between 2 and 8 is :
5+9+4+2+4
×
39
24
100 39
× 100 = 61·45
Example 3·4. In a survey, it was found that 64 families bought milk in the following quantities (litres)
in a particular week.
19
16
22
9
22
12
39
19
14
23
6
24
16
18
7
17
20
25
28
18
10
24
20
21
10
7
18
28
24
20
14
23
25
34
22
5
33
23
26
29
13
36
11
26
11
37
30
13
8
15
22
21
32
21
31
17
16
23
12
9
15
27
17
21
Using Sturges’ rule, convert the above data into a frequency distribution by ‘Inclusive Method’.
Solution. Here the total frequency is N = 64. By Sturges’ rule, the number of classes (k) is given by :
k = 1 + 3·322 log 10 64 = 1 + 3·322 × 1·8062 = 1 + 6·0002 = 7
Range = Maximum value – Minimum value = 39 – 5 = 34
3·16
BUSINESS STATISTICS
TABLE 3·15
Hence, the magnitude (i) of the class is given by
FREQUENCY DISTRIBUTION
Range
34
i = Number of classes (k) = 7 = 4·857 –~ 5.
OF THE MILK PER WEEK AMONG 64 FAMILIES
Hence taking the magnitude of each class
interval as 5, we shall get 7 classes. Since the lowest
value is 5, which is quite a convenient figure for
being taken as the lower limit of the first class, the
various classes by the inclusive method would be
5—9, 10—14, 15—19, 20—24, 25—29,
30—34, 35—39
Using tally marks, the required frequency
distribution is obtained and is given in the Table
3·15.
Milk quantity
(litres) (C.I.)
Tally Marks
5—9
10—14
15—19
20—24
25—29
30—34
35—39
||||
||||
||||
||||
||||
||||
|||
||
||||
|||| |||
|||| |||| |||
|||
Number of families
(f)
7
10
13
18
8
5
3
Total 64
Example 3·5. A college management wanted to give scholarships to B. Com. students securing 60 per
cent and above marks in the following manner :
The marks of 25 students who were eligible for
scholarship are given below :
74, 62, 84, 72, 61, 83, 72, 81, 64,
71, 63, 61, 60, 67, 74, 66, 64, 79,
73, 75, 76, 69, 68, 78 and 67.
Calculate the monthly scholarship paid to the
students.
Percentage of
Marks
60—65
65—70
70—75
75—80
80—85
Monthly Scholarship
In Rs.
250
300
350
400
450
Solution. As we are given the amount of scholarships according to the percentage of marks of the
students within classes 60—65, 65—70,…, 80—85, we shall convert the given distribution of marks into
frequency distribution with these classes as obtained in Table 3·16.
TABLE 3·16
FREQUENCY DISTRIBUTION OF MARKS OF 25 STUDENTS
Percentage of marks
Tally Marks
No. of Students (f)
Scholarship (in Rs.) (X)
Total Amount (f X)
60—65
|||| ||
7
250
1750
65—70
||||
5
300
1500
70—75
|||| |
6
350
2100
75—80
||||
4
400
1600
80—85
|||
3
450
1350
Total
∑f = 25
∑fX = 8,300
Total monthly scholarship paid to the students is : ∑ f X = Rs. 8,300.
Example 3·6. If the class mid-points in a frequency distribution of age of a group of persons are
25, 32, 39, 46, 53 and 60, find :
(a) the size of the class interval,
(b) the class boundaries, and
(c) the class limits, assuming that the age quoted is the age completed last birthday.
Solution. (a) The size (i) of the class interval is given by :
i = Difference between the mid-values of any two consecutive classes
=7
[Since 32 – 25 = 39 – 32 = … = 60 – 53 = 7]
3·17
CLASSIFICATION AND TABULATION
(b) Since the magnitude of the class is 7 and the mid-values of the
classes are 25, 32,…, 60, the corresponding class boundaries for
different classes are obtained on adding (for upper class boundaries)
and subtracting (for lower class boundaries) half the magnitude of the
class interval viz., (7/2) = 3·5, from the mid-value respectively. For
example the class boundaries for the first class will be (25 – 3·5,
25 + 3·5) i.e., (21·5, 28·5) ; for the second class will be (32 – 3·5, 32 +
3·5) i.e., (28·5, 35·5) and so on. Thus the various classes (Eclusive
Type) with class boundaries are as given in Table 3·17.
TABLE 3·17
Class
21·5—28·5
28·5—35·5
35·5—42·5
42·5—49·5
49·5—56·5
56·5—63·5
Mid-value
25
32
39
46
53
60
TABLE 3·18
(c) Assuming the age quoted (X) is the age completed on last
birthday then X will be a discrete variable which can take only
integral values. Hence the given distribution can be expressed in an
‘inclusive type’ of classes with class interval of magnitude 7, as
given in Table 3·18.
(For details see Remark 2 § 3·4·4).
Age (on last birthday)
22—28
29—35
36—42
43—49
50—56
57—63
Mid-po int
25
32
39
46
53
60
Example 3·7. The following table shows the distribution of the life time of 350 radio tubes.
Lifetime (in hours) :
Number of tubes
:
300—400
6
400—500
18
500—600
73
600—700
165
700—800
62
800—900
22
900—1000
4
Stating clearly the assumptions involved, obtain the percentage of tubes that have life time:
(a) Greater than 760 hours ; (b) Between 650 and 850 hours; and (c) Less than 530 hours.
Solution. Under the assumption that the class frequencies are uniformly distributed within the
corresponding classes, we obtain by simple interpolation technique :
(a) Number of tubes with the lifetime over 760 hours
=4+2 +
(
62
100 ×
)
40 = 26 + 24·8 = 50·8 –~ 51, since number of tubes cannot be fractional.
51
× 100 = 14·57.
350
(b) Number of tubes with lifetime over 650 hours
Hence, required percentage of tubes =
= 4 + 22 + 62 + 165
~ 171
100 × 50 = 88 + 82·5 = 170·5 –
Number of tubes with lifetime over 850 hours = 4 +
(
22
100 ×
)
50 = 15
Hence the number of tubes with lifetime between 650 hours and 850 hours is 171 – 15 = 156.
The required percentage of tubes = 156
350 × 100 = 44·57
73
(c) Number of tubes with life less than 530 hours = 6 + 18 + 100
× 30 = 6 + 18 + 21·9 = 45·9 –~ 46
46
Hence required percentage of tubes = 350
× 100 = 13·14.
3·5. CUMULATIVE FREQUENCY DISTRIBUTION
A frequency distribution simply tells us how frequently a particular value of the variable (class) is
occurring. However, if we want to know the total number of observations getting a value ‘less than’ or
‘more than’ a particular value of the variable (class), this frequency table fails to furnish the information as
such. This information can be obtained very conveniently from the ‘cumulative frequency distribution’,
which is a modification of the given frequency distribution and is obtained on successively adding
the frequencies of the values of the variable (or classes) according to a certain law. The frequencies so
obtained are called the cumulative frequencies abbreviated as c.f. The laws used are of ‘less than’ and ‘
more than’ type giving rise ‘less than cumulative frequency distribution’ and ‘more than cumulative
3·18
BUSINESS STATISTICS
frequency distribution’. We shall explain the construction of such distributions by means of a numerical
illustration.
3·5·1. Less Than Cumulative Frequency. Let us consider the distribution of marks of 70 students in a
test as given in Table 3·19.
TABLE 3·19
Less than cumulative frequency for any value of the variable (or
class) is obtained on adding successively the frequencies of all the
Marks
No. of students
previous values (or classes), including the frequency of variable
30—35
5
(class) against which the totals are written, provided the values
35—40
10
(classes) are arranged in ascending order of magnitude. For instance,
40—45
15
in the above illustration, the total number of students with marks less
45—50
30
than, say, 40 is 5 + 10 = 15; ‘less than 50’ is the sum of all the
50—55
5
previous frequencies upto and including the class 45—50 i.e.,
55—60
5
5 + 10 + 15 + 30 = 60 and so on. The final distribution is given in
Total
70
Table 3·20.
TABLE 3·20
‘LESS THAN’ CUMULATIVE FREQUENCY
DISTRIBUTION OF MARKS OF 70 STUDENTS
Marks
30—35
35—40
40—45
45—50
50—55
55—60
Frequency (f)
5
10
15
30
5
5
‘Less than’ c.f.
5
5 + 10 = 15
15 + 15 = 30
30 + 30 = 60
60 + 5 = 65
65 + 5 = 70
TABLE 3·20 (a)
LESS THAN c. f. DISTRIBUTION
Marks
Less than 30
” ” 35
” ” 40
” ” 45
” ” 50
” ” 55
” ” 60
Frequency
0
5
15
30
60
65
70
The ‘less than’ cumulative frequency distribution of Table 3·20 can also be written as given in
Table 3·20(a).
3·5·2. More Than Cumulative Frequency. The ‘more than cumulative frequency’ is obtained
similarly by finding the cumulative totals of frequencies starting from the highest value of the variable
(class) to the lowest value (class). Thus in the above illustration the number of students with marks ‘more
than 50’ is 5 + 5 = 10, and ‘more than 40’ is 15 + 30 + 5 + 5 = 55 and so on. The complete ‘more than’ type
cumulative frequency distribution for this data is given in Table 3·21.
TABLE 3·21
‘MORE THAN’ CUMULATIVE FREQUENCY
DISTRIBUTION OF MARKS OF 70 STUDENTS
Marks
30—35
35—40
40—45
45—50
50—55
55—60
Frequency
(f)
5
10
15
30
5
5
‘More than’ cumulative
frequency (c.f.)
65 + 5 = 70
55 + 10 = 65
40 + 15 = 55
10 + 30 = 40
5 + 5 = 10
5
TABLE 3·21 (a)
MORE THAN FREQUENCY
DISTRIBUTION
Marks
More than
”
”
”
”
”
”
”
”
”
”
”
”
No. of students
30
35
40
45
50
55
60
70
65
55
40
10
5
0
The ‘more than’ c.f. distribution of Table 3·21 can also be expressed as given in Table 3·21 (a).
Remarks 1. In fact ‘less than’ and ‘more than’ words also include the equality sign i.e., ‘less than a
given value’ means ‘less than or equal to that value’ and ‘more than a given value’ means ‘more than or
equal to that value’.
3·19
CLASSIFICATION AND TABULATION
2. Cumulative frequency distribution is of particular importance in the computation of median,
quartiles and other partition values of a given frequency distribution. [For details See Chapter 5—
Averages].
3. In ‘less than’ cumulative frequency distribution, the c.f. refers to the upper limit of the corresponding
class and in ‘more than’ cumulative frequency distribution, the c.f. refers to the lower limit of the
corresponding class.
Example 3·8. Convert the following distribution into ‘more than’ frequency distribution.
Weekly wages (less than ’00 Rs.)
Number of workers
:
:
20
41
40
92
60
156
80
194
100
201
Solution. Here we are given, ‘less than’ cumulative frequency distribution. To obtain the ‘more than’
cumulative frequency distribution, we shall first convert it into continuous frequency distribution as shown
in Table 3·22.
TABLE 3·22
MORE THAN
c.f. DISTRIBUTION
Weekly wages
(in ’00 Rs.)
0—20
20—40
40—60
60—80
80—100
No. of workers
(f)
41
92 – 41 = 51
156 – 92 = 64
194 – 156 = 38
201 – 194 = 7
TABLE 3·22 (a)
‘MORE THAN’ FREQUENCY
DISTRIBUTION
‘More than’
c.f.
160 + 41 = 201
109 + 51 = 160
45 + 64 = 109
7 + 38 = 45
7
Weekly wages more than
(’00 Rs.)
No. of workers
0
20
40
60
80
100
201
160
109
45
7
0
From Table 3·22, we obtain the ‘more than’ frequency distribution as given in Table 3·22 (a).
Example 3·9. The credit office of a departmental store gave the following statements for payment due
to 40 customers. Construct a frequency table of the balances due taking the class intervals as Rs. 50 and
under Rs. 200, Rs. 200 and under Rs. 350, etc. Also find the percentage cumulative frequencies and
interpret these values.
Balances due in Rs.
337,
570,
99,
759,
487,
352,
115,
60,
521,
95
563,
399,
625,
215,
360,
178,
827,
301,
501,
199
110,
501,
201,
99,
637,
328,
539,
150,
417,
250
451,
595,
422,
344,
186,
681,
397,
790,
272,
514
Solution. Taking the class intervals as 50—200, 200—350, … …, and using tally marks, we obtain the
following distribution of the balance due (in Rs.) from 40 customers.
TABLE 3·23
FREQUENCY TABLE OF BALANCE DUE (IN RUPEES) TO 40 CUSTOMERS
Balance due (in Rs.)
Tally Marks
No. of customers (f)
Less than c.f.
10
Percentage c.f.*
50—200
||||
||||
10
200—350
||||
|||
8
10 + 8 = 18
45·0
350—500
||||
|||
8
18 + 8 = 26
65·0
500—650
||||
||||
10
26 + 10 = 36
90·0
650—800
|||
3
36 + 3 = 39
97·5
800—950
|
1
39 + 1 = 40
100·0
Total
c.f.
* Percentage c.f. = Less than
× 100
N
N = ∑f = 40
25·0
3·20
BUSINESS STATISTICS
The last column of the percentage cumulative frequencies shows that 25% of the customers have to pay
less than Rs. 200, 45% of customers have to pay less than Rs. 350 ; 65% of the customers have to pay less
than Rs. 500 ; 90% of the customers have to pay less than Rs. 650 ; 97·5% of the customers have to pay less
than Rs. 800 and the balance due is less than Rs. 950 from each of the 40 customers i.e., no customer has to
pay more than Rs. 950.
3·6. BIVARIATE FREQUENCY DISTRIBUTION
So far our study was confined to frequency distribution of a single variable only. Such frequency
distributions are also called univariate frequency distributions. Quite often we are interested in
simultaneous study of two variables for the same population. This amounts to classifying the given
population w.r.t. two bases or criteria simultaneously. For example, we may study the weights and heights
of a group of individuals, the marks obtained by a group of individuals on two different tests or subjects,
income and expenditure of a group of individuals, ages of husbands and wives for a group of couples, etc.
The data so obtained as a result of this cross classification give rise to the so-called bivariate frequency
distribution and it can be summarised in the form of two-way table called the bivariate frequency table or
commonly called the correlation table. Here also the values of each variable are grouped into various
classes (not necessarily the same for each variable) keeping in view the same considerations of
classification as for a univariate distribution. If the data corresponding to one variable, say, X is grouped
into m classes and the data corresponding to the other variable, say, Y is grouped into n classes then the
bivariate table will consist of m × n cells. By going through the different pairs of the values (x, y) of the
variables and using tally marks we can find the frequency for each cell and thus obtain the bivariate
frequency table. The format of a bivariate frequency table is given in Table 3.24.
TABLE 3·24
BIVARIATE FREQUENCY TABLE
X Series
Y Series
Classes
→
↓
Total of
frequencies
of Y
Mid Points
x1
x2
...
x
...
xm
Mid Points
Classes
y1
y2
y
f (x, y)
fy
..
.
yn
Total of
frequencies
of X
fx
Total
∑ fx =∑ fy = N
Here f (x, y) is the frequency of the pair (x, y).
Remarks 1. The bivariate frequency table gives a general visual picture of the relationship between the
two variables under consideration. However, a quantitative measure of the linear relationship between the
variables is given by the correlation coefficient (See Chapter 8, Correlation Analysis).
2. Marginal Distributions of X and Y. The frequency distribution of the values of the variable X
together with their frequency totals as given by fx in the above table is called the marginal frequency
distribution of X. Similarly, the frequency distribution of the values of the variable Y together with the total
frequencies fy in the above table, is known as the marginal frequency distribution of Y.
3·21
CLASSIFICATION AND TABULATION
3. Conditional Distributions of X and Y. The conditional frequency distribution of X for a given
value of Y is obtained by the values of X together with their frequencies corresponding to the fixed values
of Y. Similarly, we may obtain the conditional frequency distribution of Y for given values of X.
We shall now explain the technique of constructing bivariate frequency table and obtaining the
marginal and conditional distributions of X and Y by means of numerical illustrations.
Example 3·10. (a) Prepare a bivariate frequency distribution for the following data for 20 students :
Marks in Law
10
11
10
11
11
14
12
12
13
10
Marks in Statistics
29
21
22
21
23
23
22
21
24
23
Marks in Law
13
12
11
12
10
14
14
12
13
10
Marks in Statistics
24
23
22
23
22
22
24
20
24
23
(b) Also obtain the marginal frequency distributions of the marks in Law and marks in Statistics and
the conditional frequency distribution of marks in Law when marks in Statistics are 23 and the conditional
distribution of marks in Statistics when marks in Law are 12.
Solution. (a) Let us denote the marks in Law by the variable X and the marks in Statistics by the
variable Y. Then X takes the values from 10 to 14 i.e., 5 values in all, and Y takes the values from 20 to 24
i.e., 5 values in all. Thus the two-way table will consist of 5 × 5 = 25 cells.
To prepare the bivariate frequency table, we observe that the first student gets 10 marks in Law and 20
marks in Statistics. Therefore, we put a tally mark in the cell where the column corresponding to X = 10
intersects the row corresponding to Y = 20. Proceeding similarly we put tally marks for each pair of values
(x, y) for all the 20 candidates. The total frequency for each cell is given in small brackets ( ), after the tally
marks. Now count all the frequencies in each row and write at extreme right column. Similarly count all the
frequencies in each column and write at the bottom row. The bivariate frequency distribution so obtained is
given in Table 3·25.
TABLE 3·25
BIVARIATE FREQUENCY TABLE SHOWING MARKS
OF 20 STUDENTS IN LAW AND STATISTICS
Marks in
Law
(X) →
Marks in
Statistics
(Y) ↓
20
10
11
| (1)
21
12
13
|| (2)
| (1)
|| (2)
| (1)
| (1)
23
| (1)
5
3
|| (2)
| (1)
|| (2)
24
5
Total
(fy)
2
| (1)
22
Total (fx )
14
4
5
| (1)
6
||| (3)
| (1)
4
3
3
20
(b) The marginal frequency distributions of X and Y are given in Table 3·25(a).
TABLE 3·25 (a)
MARGINAL DISTRIBUTIONS
Marginal Distribution
of X
X
10
11
12
13
14
Total
f
5
4
5
3
3
20
Marginal Distribution
of Y
Y
20
21
22
23
24
Total
f
2
3
5
6
4
20
TABLE 3·25 (b)
CONDITIONAL DISTRIBUTIONS
Conditional Distribution
of X when Y = 23
X
10
11
12
13
14
Total
Frequency
2
1
2
0
1
6
Conditional Distribution
of Y when X = 12
Y
20
21
22
23
24
Total
Frequency
1
1
1
2
0
5
3·22
BUSINESS STATISTICS
The conditional distributions of marks in Law (X) when marks in Statistics i.e., Y = 23 and the
conditional distribution of marks in Statistics (Y) when marks in Law, i.e., X = 12, are given in
Table. 3·25(b).
Example 3·11. Following figures give the ages in years of newly married husbands and wives.
Represent the data by a frequency distribution.
Age of Husband : 24
26
27
25
28
24
27
28
25
26
Age of Wife
: 17
18
19
17
20
18
18
19
18
19
Age of Husband : 25
26
27
25
27
26
25
26
26
26
Age of Wife
: 17
18
19
19
20
19
17
20
17
18
[Delhi Univ. B.Com. (Hons.), 1975]
Solution. Let us denote the age (in years) of the husbands by the variable X and the age (in years) of
wives by the variable Y. Then we observe that the variable X takes the values from 24 to 28 and Y takes the
values from 17 to 20. Proceeding exactly as in Example 3·10, we obtain the bivariate frequency distribution
given in Table 3·26.
TABLE 3·26
FREQUENCY DISTRIBUTION OF THE AGES (IN YEARS)
OF NEWLY MARRIED HUSBANDS AND WIVES
Age of
Husband
Age of
(X) →
Wife
(Y) ↓
24
25
26
27
17
18
19
20
| (1)
| (1)
||| (3)
| (1)
| (1)
| (1)
||| (3)
|| (2)
| (1)
| (1)
|| (2)
| (1)
| (1)
| (1)
5
6
6
3
Total (fx )
2
4
2
20
5
7
28
Total
(fy)
EXERCISE 3·1.
1. (a) What do you mean by classification of data ? Discuss in brief the modes of classification.
[Delhi Univ. B.Com. (Pass), 1996]
(b) Briefly explain the principles of Classification.
[Delhi Univ. B.Com. (Pass), 2000]
(c) What do you understand by classification of data ? What are its objectives ?
(d) What are different types of classification ? Illustrate by suitable examples.
2. “Classification is the process of arranging things (either actually or notionally) in groups or classes according to
their resemblances and affinities giving expression to the unity of attributes that may subsist amongst a diversity of
individuals”.
Elucidate the above statement.
3. (a) What is meant by ‘classification’? State its important objectives. Briefly explain the different methods of
classifying statistical data.
[C.A. (Foundation), June 1993]
(b) What are the purposes of classification of data ? State the primary rules to be observed in classification.
[C.A. (Foundation), Nov. 1995]
4. (a) What are the advantages of data classification? What primary rules should ordinarily be followed for
classification ?
[C.A. (Foundation), Nov. 1997]
(b) What are different kinds of classification ? State, how the classification of data is useful.
[C.A. (Foundation), Nov. 2000]
(c) State the principles underlying classification of data.
5. (a) What are grouped and ungrouped frequency distributions ? What are their uses ? What are the considerations
that one has to bear in mind while forming the frequency distribution ?
(b) Discuss the problems in the construction of a frequency distribution from raw data, with particular reference to
the choice of number of classes and the class limits.
CLASSIFICATION AND TABULATION
3·23
6. (a) What are the principles governing the choice of :
(i) Number of class intervals.
(ii) The length of the class interval.
(iii) The mid point of class interval ?
(b) What are the general rules of forming a frequency distribution with particular reference to the choice of classinterval and number of classes ? Illustrate with examples.
[Calicut Univ., B.Com., 1999]
7. What do you mean by an inclusive series ? How can an inclusive series be converted to an exclusive series ?
Illustrate with the help of an example.
[Delhi Univ. B.Com. (Pass), 2002]
8. Prepare a frequency distribution from the following figures relating to bonus paid to factory workers :
BONUS PAID TO WORKERS (in Rs.)
86
62
58
73
101
90
84
90
76
61
84
63
56
72
102
56
83
92
87
60
83
69
57
71
103
57
87
93
88
59
76
70
54
70
104
58
88
94
89
57
74
86
55
60
105
59
89
84
90
74
67
84
82
70
60
60
90
81
91
76
60
83
78
80
65
61
96
82
93
102
60
91
76
90
70
63
94
83
94
101
70
100
74
101
80
67
92
85
96
92
Take a class-interval of 5.
Ans. Frequencies of classes 50—54, 55—59,…, 100—104, 105—109 are respectively 1, 10, 11, 4, 11, 5, 13, 9,
15, 2, 8 and 1.
9. Describe in brief the practical problems of frequency distribution in classification of data on a variable
according to class intervals, giving the definition of frequency distribution.
[C.A. (Foundation), Nov. 2001]
10. (a) For the following raw data prepare a frequency distribution with the starting class as 5—9 and all classes
with the same width 5.
Marks in English
12
36
40
16
10
10
19
20
28
30
19
27
15
21
33
45
7
19
20
26
26
37
6
5
20
30
37
17
11
20
Ans. Marks
: 5—9 10—14 15—19 20—24 25—29 30—34 35—39 40—44 45—49
Frequency :
3
4
6
5
4
3
3
1
1
(b) Classify the following data by taking class intervals such that their mid-values are 17, 22, 27, 32, and so on.
30
42
30
54
40
48
15
17
51
42
25
41
30
27
42
36
28
26
37
54
44
31
36
40
36
22
30
31
19
48
16
42
32
21
22
46
33
41
21
[Madurai-Kamaraj Univ. B.Com., 1995]
Ans. 15—19
20—24
25—29
30—34
35—39
40—44
45—49
50—54
4
4
4
8
4
9
3
3
11. (a) The following are the weights in kilograms of a group of 55 students.
42
74
40
60
82
115
41
61
75
83
63
53
110
76
84
50
67
78
77
63
65
95
68
69
104
80
79
79
54
73
59
81
100
66
49
77
90
84
76
42
64
69
70
80
72
50
79
52
103
96
51
86
78
94
71
Prepare a frequency table taking the magnitude of each class-interval as 10 kg. and the first class-interval as equal
to 40 and less than 50.
Ans. Frequencies of classes 40—50, 50—60,…,110—120 are 5, 7, 11, 15, 8, 4, 3, 2 respectively.
(b) Prepare a statistical table from the following data taking the class width as 7 by inclusive method.
24
26
28
32
37
5
1
7
9
11
15
13
14
18
29
31
32
6
4
2
9
18
27
36
3
9
15
21
27
33
4
8
12
16
20
5
10
3
8
1
6
4
9
2
7
12
18
27
23
21
29
22
15
17
28
Ans. Frequencies of classes 1—7, 8—14, 15—21,…, 36—42 are respectively 15, 12, 11, 9, 6, 2.
3·24
BUSINESS STATISTICS
12. Using Sturges’ Rule k = 1 + 3·322 log N, where k is the number of class intervals, N is the total number of
observations, classify in equal intervals, the following data of hours worked by 50 piece rate workers for a period of a
month in a certain factory :
110, 175, 161, 157, 155, 108, 164, 128, 114, 178, 165, 133, 195, 151,
71,
94,
97,
42,
30,
62, 138, 156, 167, 124, 164, 146, 116, 149, 104, 141, 103, 150, 162, 149,
79, 113,
69, 121,
93, 143, 140, 144, 187, 184, 197,
87,
40, 122, 203, 148.
Ans. Using Sturges’ rule we get k (No. of classes) = 7, Magnitude of class = (Range/k) = (174/7) ~
– 25.
Classes are 30—55, 55—80, 80—105,…,180—205. The corresponding frequencies are 3, 4, 6, 9, 12, 11, 5.
13. Two dice are thrown at random. Obtain the frequency distribution of the sum of the numbers which appear on
them.
Hint and Ans. Total possible pairs of numbers on the two dice are as given below :
(1, 1), (1, 2),…, (1, 6) ; (2, 1), (2, 2),…, (2, 6) ; … …, (6, 1), (6, 2_,…, (6, 6)
If X denotes the sum of the numbers on the two dice then X is a discrete variable which can take the values 2, 3,…, 12.
Sum (X) :
2
3
4
5
6
7
8
9
10
11
12
Frequency :
1
2
3
4
5
6
5
4
3
2
1
14. If the class mid-points in a frequency distribution of a group of persons are : 125, 132, 139, 146, 153, 160, 167,
174, 181 pounds, find (i) size of the class intervals, (ii) the class boundaries, and (iii) the class limits,
assuming that the weights are measured to the nearest pound.
[Delhi Univ. B.Com. (Hons.), 2007]
Ans.
(i) 7
(ii) 121·5—128·5, 128·5—135·5,…,177·5—184·5
(iii) 122—128, 129—135,…, 178—184.
15. With the help of suitable examples, distinguish between :
(i) Continuous and Discrete variable.
(ii) Exclusive and Inclusive class intervals.
(iii) ‘More than’ and ‘Less than’ frequency tables. (iv) Simple and Bivariate frequency tables.
16. What do you mean by cumulative frequency (c.f.) distribution ; ‘More than’ and ‘Less than’ type c.f.
distribution. Illustrate by an example.
17. The weekly observations on cost of living index in a certain city for the year 2000-01 are given below :
Cost of living index :
140—150 150—160 160—170 170—180
180—190
190—200
No. of workers :
5
10
20
9
6
2
Prepare ‘less than’ and ‘more than’ cumulative frequency distributions.
18. (a) Convert the following into an ordinary frequency distribution :
5 students get less than 3 marks ;
12 students get less than 6 marks;
25 students get less than 9 marks ;
33 students get less than 12 marks.
[Delhi Univ. B.Com. (Pass), 2001]
Ans.
0—3
3—6
6—9
9—12
5
7
13
8
(b) Following is a cumulative frequency table showing the number of packages and the number of times a given
number of packages was received by a post office in 60 days :
No. of packages below
:
10
20
30
40
50
60
No. of times received in 60 days :
17
22
29
37
50
60
Obtain the frequency table from it. Also prepare ‘more than’ cumulative frequency table.
19. (a) What is the difference between continuous and discrete variables ?
(b) Are the following variables discrete or continuous ? Give your answer with reason.
(i) Age on last birthday.
;
(ii) Temperature of the patient.
(iii) Length of a room.
;
(iv) Number of shareholders in a company.
Ans.
(ii) and (iii) continuous ; (i) and (iv) discrete.
(c) State with reasons which of the following represent discrete data and which represent continuous data :
(i) Number of table fans sold each day at a Departmental Store.
(ii) Temperature recorded every half an hour of a patient in a hospital.
(iii) Life of television tubes produced by Electronics Ltd.
(iv) Yearly income of school teachers.
(v) Lengths of 1,000 bolts produced in a factory.
3·25
CLASSIFICATION AND TABULATION
20. Complete the table showing the frequencies with which words of different number of letters occur in the
extract reproduced below (omitting punctuation marks) treating as the variable the number of letters in each word :
“Her eyes were blue ; blue as autumn distance—blue as the blue we see between the retreating mouldings of hills
and woody slopes on a sunny September morning : a misty and shady blue, that had no beginning or surface, and was
looked into rather than at.”
Ans.
X:
1
2
3
4
5
6
7
8
9
10
f:
2
8
9
10
5
4
3
1
3
1
21. Prepare a frequency distribution of the words in the following extract according to their length (number of
letters) omitting punctuation marks. Also give (i) the number of words with 7 letters or less ; (ii) the proportion of
words with 5 letters or more ; (iii) the percentage of words with not less than 4 and not more than 7 letters.
“Success in the examination confers no absolute right to appointment, unless Government is satisfied, after such
enquiry as may be considered necessary, that the candidate is suitable in all respects for appointment to the public
service.”
Ans.
X:
2
3
4
5
6
7
8
9
10
11
f:
9
6
2
2
2
4
3
3
2
3
(i) 25,
(ii) (19/36) = 0·5278,
(iii) (10/36) × 100 = 27·28.
22. A company wants to pay daily bonus to its employees. The bonus is to be paid us under :
Daily Salary (Rs.) :
100—200
200—300
300—400
400—500
500—600
Daily Bonus (Rs.) :
10
20
30
40
50
Actual daily salaries of the employees, in Rupees, are as under :
175,
225,
375,
478,
525,
650,
570,
451,
382,
280
375,
465,
530,
480,
320,
515,
225,
345,
471,
450
Find out the total daily bonus paid to the employees.
600—700
60
Ans. Total daily bonus paid to the employees = Rs. 720.
23. From the following data construct a bivariate frequency distribution :
Age of husbands
(in years) (x)
28
26
27
25
28
Ans.
x:
fx :
Age of wives
(in years) (y)
22
21
21
20
22
25
26
2
4
Age of husbands
(in years) (x)
28
27
27
26
27
27
28
6
3
Age of wives
Age of husbands
(in years) (y)
(in years) (x)
21
27
21
26
20
25
20
26
19
27
y:
19
20
21
fy :
3
4
6
Age of wives
(in years) (y)
21
19
19
20
21
22
2
24. The data given below relates to the heights and weights of 20 persons. You are required to form a two-way
frequency table with class 62″ to 64″, 64″ to 66″ and so on, and 115 to 125 lbs, 125 to 135 lbs. and so on.
S. No.
1.
2.
3.
4.
5.
6.
7.
Weight
170
135
136
137
148
124
117
Height
70
65
65
64
69
63
65
S. No.
8.
9.
10.
11.
12.
13.
14.
Weight
128
143
129
163
139
122
134
Height
70
71
62
70
67
63
68
S. No.
15.
16.
17.
18.
19.
20.
Weight
140
132
120
148
129
152
Height
67
69
66
68
67
67
Ans. [Frequencies: (W) = 4, 5, 6, 3, 1, 1 ; (H) = 3, 4, 5, 4, 4.]
25. The following figures are income (x) and percentage expenditure on food (y) in 25 families. Construct a
bivariate frequency table classifying x into intervals 200—300, 300—400, …, and y into 10—15, 15—20, … .
Write down the marginal distributions of x and y and the conditional distribution of x when y lies between 15
and 20.
3·26
BUSINESS STATISTICS
x
y
x
y
x
y
x
y
x
y
550
623
310
420
600
12
14
18
16
15
225
310
640
512
690
25
26
20
18
12
680
330
425
555
325
13
25
16
15
23
202
255
492
587
643
29
27
18
21
19
689
523
317
384
400
11
12
18
17
19
26. What is meant by a bi-variate series? Taking 15 imaginary figures, construct a series of this type on discrete
pattern.
[Delhi Univ. B.Com. (Pass), 1998]
27. 30 pairs of values of two variables X and Y are given below. Form a two-way table :
X
14
20
33
25
41
18
24
29
38
45
23
32
37
19
28
Y
148 242
296 312
518
196
214
340 492
568
282 400 288 292
431
X
34
38
29
44
40
22
39
43
44
12
27
39
38
17
26
Y
440 500
512 415
514
282
481
516 598
122
200 451 387 245
413
Take class intervals of X as 10 to 20, 20 to 30 etc., and that of Y as 100 to 200, 200 to 300, etc.
[Osmania Univ. B.Com., 1996]
28. Following are the marks obtained by 24 students in English (X) and Economics (Y) in a test.
(15, 13), (0, 1), (1, 2), (3, 7), (16, 8),
(2, 9), (18, 12), (5, 9), (4, 17), (17, 16), (6, 6), (19, 18)
(14, 11), (9, 3), (8, 5), (13, 4), (10, 10), (13, 11), (11, 14), (11, 7), (12, 18), (18, 15), (9, 15), (17, 3).
Taking class-intervals as 0—4, 5—9, etc., for X and Y both, construct—
(i) Bivariate frequency table.
(ii) Marginal frequency tables of X and Y.
29. Fill in the blanks :
(i) Variables are of two kinds … … and … …
(ii) … … is the process of arranging data into groups according to their common characteristics.
(iii) In chronological classification, the data are classified on the basis of … …
(iv) … … classification means the classification of data according to location.
(v) Class-mark (mid-point) is the value lying half-way between … …
(vi) According to Sturges’ rule, the number of classes (k) is given by : k = … …
(vii) The magnitude of the class (i) is given by :
i=……
(viii) … … of data is a function very similar to that of sorting letters in a post office.
(ix) Different bases of classification of data are … …
(x) The data can be classified into … … and … … type classes.
(xi) While forming a grouped frequency distribution, the number of classes should usually be between … …
(xii) In exclusive type classes, the upper limit of the class is … …
(xiii) In the continuous classes 0—5, 5—10, 10—15, 15—20 and so on, the class 15—20 means that the
variable X takes the values … …
(xiv) Two examples of discrete variable are … … and … … and continuous variable are … … and … …
(xv) The classes in which the lower limit or the upper limit are not specified, are known as … …
(xvi) The difference between the upper and the lower limits of a class gives … … of the class.
(xvii) The number of observations in a particular class is called the … … of the class.
(xviii) If the data values are classified into the classes 0—9, 10—19, 20—29, and so on and the frequency of the
class 20—29 is 12, it means that … … .
(xix) If the mid-points of the classes are 16, 24, 32, 40, and so on, then the magnitude of the class intervals is …
….
(xx) In (xix), the class boundaries are … … .
Ans. (i) discrete, continuous, (ii) classification. (iii) time. (iv) geographical. (v) the upper and the lower limits of
the class. (vi) k = 1 + 3·322 log10 N ; N is total frequency. (vii) i = (upper limit – lower limit) of the class.
(viii) classification. (ix) geographical, chronological, qualitative and quantitative. (x) inclusive, exclusive. (xi) 5 and 15.
(xii) is not included in the class. (xiii) 15 and more but less than 20 i.e., 15 ≤ X < 20. (xiv) marks in a test, number of
CLASSIFICATION AND TABULATION
3·27
accidents; height in inches, weight in kgs. (xv) open end classes. (xvi) the width or the magnitude. (xvii) frequency.
(xviii) there are 12 observations taking values between 20 and 29, both inclusive i.e., 20 ≤ X ≤ 29. (xix) 8. (xx) 12—20,
20—28, 28—36, 36—44 and so on.
3·7. TABULATION – MEANING AND IMPORTANCE
By tabulation we mean the systematic presentation of the information contained in the data, in rows
and columns in accordance with some salient features or characteristics. Rows are horizontal arrangements
and columns are vertical arrangements. In the words of A.M. Tuttle :
“A statistical table is the logical listing of related quantitative data in vertical columns and horizontal
rows of numbers with sufficient explanatory and qualifying words, phrases and statements in the form of
titles, headings and notes to make clear the full meaning of data and their origin.”
Professor Bowley in his manual of Statistics refers to tabulation as “the intermediate process between
the accumulation of data in whatever form they are obtained, and the final reasoned account of the result
shown by the statistics.”
Tabulation is one of the most important and ingenious device of presenting the data in a condensed and
readily comprehensible form and attempts to furnish the maximum information contained in the data in the
minimum possible space, without sacrificing the quality and usefulness of the data. It is an intermediate
process between the collection of the data on one hand and statistical analysis on the other hand. In fact,
tabulation is the final stage in collection and compilation of the data and forms the gateway for further
statistical analysis and interpretations. Tabulation makes the data comprehensible and facilitates
comparisons (by classifying data into suitable groups), and the work of further statistical analysis,
averaging, correlation, etc. It makes the data suitable for further Diagrammatic and Graphic representation.
If the information contained in the data is expressed as a running text using language paragraphs, it is
quite time-consuming to comprehend it because in order to understand every minute details of the text, one
has to go through all the paragraphs ; which usually contain very large amount of repetitions. Tabulation
overcomes the drawback of the repetition of explanatory phrases and headings and presents the data in a
neat, readily comprehensible and true perspective, thus highlighting the significant and relevant details and
information. Tabulated data have attractive get up and leave a lasting impression on the mind as compared
to the data in the textual form. Tabulation also facilitates the detection of the errors and the omissions in the
data. Tabulation enables us to draw the attention of the observer to specific items by means of comparisons,
emphasis and arrangement of the layout.
No hard and fast rules can be laid down for tabulating the statistical data. To prepare a first class table
one must have a clear idea about the facts to be presented and stressed, the points on which emphasis is to
be laid and familiarity with technique of preparation of the table. The arrangement of data tabulation
requires considerable thought to ensure showing the relationship between the data of one or more series, as
well as the significance of all the figures given in the classification adopted. The facts, comparisons and
contrasts, and emphasis vary from one table to another table. Accordingly a good table (the requirements of
which are given below) can only be obtained through the skill, expertise, experience and common sense of
the tabulator, keeping in view the nature, scope and objectives of the enquiry. This bears testimony to the
following words of A.L. Bowley :
“In the tabulation of the data common sense is the chief requisite and experience is the chief teacher.”
3·7·1. Parts of a Table. The various parts of a table vary from problem to problem depending upon the
nature of the data and the purpose of the investigation. However, the following are a must in a good
statistical table :
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
Table number
Title
Head notes or Prefatory notes
Captions and Stubs
Body of the table
Foot-note
Source note
3·28
BUSINESS STATISTICS
1. Table Number. If a book or an article or a report contains more than one table then all the tables
should be numbered in a logical sequence for proper identification and easy and ready reference for future.
The table number may be placed at the top of the table either in the centre above the title or in the side of
the title.
2. Title. Every table must be given a suitable title, which usually appears at the top of the table (below
the table number or next to the table number). A title is meant to describe in brief and concise form the
contents of the table and should be self-explanatory. It should precisely describe the nature of the data
(criteria of classification, if any) ; the place (i.e., the geographical or political region or area to which the
data relate) ; the time (i.e., period to which the data relate) and the source of the data. The title should be
brief but not an incomplete one and not at the cost of clarity. It should be un-ambiguous and properly
worded and punctuated. Sometimes it becomes desirable to use long titles for the sake of clarity. In such a
situation a ‘catch title’ may be given above the ‘main title’. Of all the parts of the table, title should be most
prominently lettered.
3. Head Notes (or Prefatory Notes). If need be, head note is given just below the title in a prominent
type usually centred and enclosed in brackets for further description of the contents of the table. It is a sort
of a supplement to the title and provides an explanation concerning the entire table or its major parts-like
captions or stubs. For instance, the units of measurements are usually expressed as head such as ‘in
hectares’, ‘in millions’, ‘in quintals’, ‘in Rupees’, etc.
4. Captions and Stubs. Captions are the headings or designations for vertical columns and stubs are
the headings or designations for the horizontal rows. They should be brief, concise and self-explanatory.
Captions are usually written in the middle of the columns in small letters to economise space. If the same
unit is used for all the entries in the table then it may be given as a head note along with the title. However,
if the items in different columns or rows are measured or expressed in different units, then the
corresponding units should also be indicated in the columns or rows. Relative units like ratios, percentages,
etc., if any, should also be specified in the respective rows or columns. For instance, the columns may
constitute the population (in millions) of different countries and rows may indicate the different periods
(years).
Quite often two or more columns or rows corresponding to similar classifications (or with same
headings) may be grouped together under a common heading to avoid repetitions and may be given what
are called sub-captions or sub-stubs. It is also desirable to number each column and row for reference and
to facilitate comparisons.
5. Body of the Table. The arrangement of the data according to the descriptions given in the captions
(columns) and stubs (rows) forms the body of the table. It contains the numerical information which is to be
presented to the readers and forms the most important part of the table. Undesirable and irrelevant (to the
enquiry) information should be avoided. To increase the usefulness of the table, totals must be given for
each separate class/category immediately below the columns or against the rows. In addition, the grand
totals for all the classes for rows/columns should also be given.
6. Foot Note. When some characteristic or feature or item of the table has not been adequately
explained and needs further elaboration or when some additional or extra information is required for its
complete description, foot-notes are used for this purpose. As the name suggests, footnotes, if any, are
placed at the bottom of the table directly below the body of the table. Foot-notes may be attached to the
title, captions, stubs or any part of the body of the table. Foot-notes are identified by the symbols *, **, ***,
€€†, @, etc.
7. Source Note. If the source of the table is not explicitly contained in the title, it must be given at the
bottom of the table, below the footnote, if any. The source note is required if the secondary data are used. If
the data are taken from a research journal or periodical, then the source note should give the name of the
journal or periodical along with the date of publication, its volume number, table number (if any), page
number, etc., so that anybody who uses this data may satisfy himself, (if need be), about the accuracy of the
figures given in the table by referring to the original source. Source note will also enable the user to decide
about the reliability of the data since to the learned users of Statistics the reputations of the sources may
vary greatly from one agency to another.
3·29
CLASSIFICATION AND TABULATION
The format of a blank table is given in Table 3·27.
TABLE 3·27. FORMAT OF A BLANK TABLE
TITLE
[Head Note or Prefatory Note (if any)]
Caption
Stub
Heading
↓
Sub-Heads
Column
Head
Column
Head
Sub-Heads
Column
Head
Column
Head
Total
Column
Head
Body
Total
Foot Note :
Source Note :
Remarks 1. A table should be so designed that it is neither too long and narrow nor too short and
broad. It should be of reasonable size adjusted to the space at our disposal and should have an attractive get
up. If the data are very large they should not be crowded in a single table which would become unwieldy
and difficult to comprehend. In such a situation it is desirable to split the large table into a number of tables
of reasonable size and shape. Each table should be complete in itself.
2. If the figures corresponding to certain items in the table are not available due to certain reasons, then
the gaps arising therefrom should be filled by writing N.A. which is used as an abbreviation for ‘not
available’.
3·7·2. Requisites of a Good Table. As pointed out earlier, no hard and fast rules can be laid down for
preparing a statistical table. Preparation of a good statistical table is a specialised job and requires great
skill, experience and common sense on the part of the tabulator. However, commensurate with the
objectives and scope of the enquiry, the following points may be borne in mind while preparing a good
statistical table.
(i) The table should be simple and compact so that it is readily comprehensible. It should be free from
all sorts of overlappings and ambiguities.
(ii) The classification in the table should be so arranged as to focus attention on the main comparisons
and exhibit the relationship between various related items and facilitate statistical analysis. It should
highlight the relevant and desired information needed for further statistical investigation and emphasise the
important points in a compact and concise way. Different modes of lettering (in italics, bold or antique
type, capital letters or small letters of the alphabet, etc.), may be used to distinguish points of special
emphasis.
(iii) A table should be complete and self-explanatory. It should have a suitable title, head note (if
necessary), captions and stubs, and footnote (if necessary). If the data are secondary, the source note should
also be given. [For details see § 3·7·1]. The use of dash (—) and ditto marks (,,) should be avoided. Only
accepted common abbreviations should be used.
(iv) A table should have an attractive get up which is appealing to the eye and the mind so that the
reader may grasp it without any strain. This necessitates special attention to the size of the table and proper
spacings of rows and columns.
3·30
BUSINESS STATISTICS
(v) Since a statistical table forms the basis for statistical analysis and computation of various statistical
measures like averages, dispersion, skewness, etc., it should be accurate and free from all sorts of errors.
This necessitates checking and re-checking of the entries in the table at each stage because even a minor
error of tabulation may lead to very fallacious conclusions and misleading interpretations of the results.
(vi) The classification of the data in the table should be in alphabetical, geographical or chronological
order or in order of magnitude or importance to facilitate comparisons.
(vii) A summary table [See § 3·7·3] should have adequate interpretative figures like totals, ratios,
percentages, averages, etc.
3·7·3. Types of Tabulation. Statistical tables are constructed in many ways. Their choice basically
depends upon :
(i) Objectives and scope of the enquiry.
(ii) Nature of the enquiry (primary or secondary).
(iii) Extent of coverage given in the enquiry.
The following diagrammatic scheme elegantly displays the various forms of tables commonly used in
practice.
Types of Tables
On the basis of
objectives or purpose
General Purpose
or Reference Table
Special Purpose
or Summary Table
On the basis of
nature of enquiry
Original
or Primary Table
Derived or
Derivative Table
On the basis of
coverage
Simple
Table
Complex
Table
General Purpose (or Reference) and Special Purpose (or Summary) Tables. General purpose
tables, which are also known as reference tables or sometimes informative tables provide a convenient way
of compiling and presenting a systematically arranged data, usually in chronological order, in a form which
is suitable for ready reference and record without any intentions of comparative studies, relationship or
significance of figures. Most of the tables prepared by government agencies e.g., the detailed tables in the
census reports, are of this kind. These tables are of repository nature and mainly designed for use by
research workers, statisticians and are generally given at the end of the report in the form of an appendix.
Examples of such tables are : age and sexwise distribution of the population of a particular region,
community or country ; payrolls of a business house ; sales orders for different products manufactured by a
concern ; the distribution of students in a university according to age, sex and the faculty they join ; and so
on.
As distinct from the general purpose or reference tables, the special purpose or summary tables (also
sometimes called interpretative tables) are of analytical nature and are prepared with the idea of making
comparative studies and studying the relationship and the significance of the figures provided by the data.
These are generally constructed to emphasise some facts or relationships pertaining to a particular or
specific purpose. In such tables interpretative figures like ratios, percentages, etc., are used in order to
facilitate comparisons. Summary tables are sometimes called derived or derivative tables (discussed below)
as they are generally derived from the general purpose tables.
Original and Derived Tables. On the basis of the nature or originality of the data, the tables may be
classified into two classes :
(i) Primary tables
(ii) Derived or Derivative tables.
In a primary table, the statistical facts are expressed in the original form. It, therefore, contains absolute
and actual figures and not rounded numbers or percentages. On the other hand, derived or derivative table is
one which contains figures and results derived from the original or primary data. It expresses the
information in terms of ratios, percentages, aggregates or statistical measures like average, dispersion,
3·31
CLASSIFICATION AND TABULATION
skewness, etc. For instance, the time series data is expressed in a primary table but a table expressing the
trend values and seasonal and cyclic variations is a derived table. In practice, mixtures of primary and
derived tables are generally used, an illustration being given below :
TABLE 3·28. LOAD CARRIED BY RAILWAYS AND ROAD TRANSPORT FOR DIFFERENT YEARS
(In Billion Tonne Km)
Year
Railways
Road Transport
1960-61
1965-66
1968-69
1973-74
1974-75
1975-76
88
117
125
122
134
148
17
34
40
65
80
81
Percentage Share
Railways
Road Transport
83·8
77·5
75·8
65·2
62·6
64·6
16·2
22·5
24·2
34·8
37·4
35·4
Simple and Complex Tables. In a simple table the data are classified w.r.t. a single characteristic and
accordingly it is also termed as one-way table. On the other hand, if the data are grouped into different
TABLE 3·29
classes w.r.t. two or more characteristics or criteria
simultaneously, then we get a complex or manifold table. In
IMPORTS FROM PRINCIPAL
COUNTRIES BY SEA, AIR AND
particular, if the data are classified w.r.t. two (three) characLAND FOR 1975-76
teristics simultaneously we get a two-way (three-way) table.
(Rupees in lakhs)
Simple Table. As already stated, a simple table furnishes
information about only one single characteristic of the data. For
Imports
Country
instance, Table 3·1 (on page 3·4) relating to agricultural output
10,167
of different countries (in kg per hectare) ; Table 3·2 (on page Australia
Canada
23,201
3·4) giving the density of population (per square kilometre) in
France
19,653
different cities of India ; Table 3·3 (on page 3·4) giving the
(West) Germany
36,996
population of India (in crores) for different years, are all simple
Japan
36,118
tables. As another illustration, the Table 3·29 giving the imports
UK
28,400
from principal countries (by sea, air and land) for the year
USA
1,28,522
1975-76 is a simple table.
USSR
30,978
Two-way Table. However, if the caption or stub is classified into two sub-groups, which means that the data are classified w.r.t. two characteristics, we get a twoway table. Thus a two-way table furnishes information about two inter-related characteristics of a particular
phenomenon. For example, the distribution of the number of students in a college w.r.t. age (1st
characteristic) and sex (2nd characteristic) gives a two-way table. As another illustration, Table 3·30 which
gives the load/distance by Railways and Road Transport for different years, is a two-way table.
TABLE 3·30. LOAD CARRIED BY RAILWAYS AND ROAD TRANSPORT FOR DIFFERENT YEARS
(In Billion Tonne Km)
Years
Railways
Road Transport
1960-61
1965-66
1968-69
1973-74
1974-75
1975-76
88
117
125
122
134
148
17
34
40
65
80
81
Three-way Tables. If the data are classified simultaneously w.r.t. three characteristics, we get a threeway table. Thus a three-way table gives us information regarding three inter-related characteristics of a
particular phenomenon. For example, the classification of a given population w.r.t. age, sex and literacy, or
the classification of the students in a university w.r.t. sex, faculty (Arts, Sciences, Commerce) and the class
(Ist year, 2nd year, 3rd year of the under-graduate courses) will give rise to three-way tables. The tables
3·32
BUSINESS STATISTICS
given in Examples 3·12 to 3·19 are three-way tables. As another illustration, the following Table 3·31
representing the distribution of population of a city according to different age-groups (say, five age groups
from 0 to 100 years), sex and literacy is a three-way table.
Row-totals
Females
Males
Sub-totals
Females
Males
Sub-totals
Males
Age Group
Females
TABLE 3·31. DISTRIBUTION OF POPULATION (IN ’000) OF A CITY
w.r.t AGE, SEX AND LITERACY
Literates
Illiterates
Total
0—20
20—40
40—60
60—80
80—100
Column
Totals
Higher Order or Manifold Tables. These tables give the information on a large number of interrelated problems or characteristics of a given phenomenon. For example, the distribution of students in a
college according to faculty, class, sex and year (Example 3·20) or the distribution of employees in a
business concern according to sex, age-groups, years and grades of salary (Example 3·21) gives rise to
manifold tables. Manifold or higher order tables are commonly used in presenting population census data.
Remark. It may be pointed out that as the order of the table goes on increasing, the table becomes
more and more difficult to comprehend and might even become confusing. In practice, in a single table
only upto three or sometimes four characteristics are represented simultaneously. If the study is confined to
more than four characteristics at a time then it is desirable to represent the data in more than one table for
depicting the relationship between different characteristics.
Example 3·12. Present the following information in a suitable tabular form, supplying the figures not
directly given :
In 1995 out of total 2000 workers in a factory, 1550 were members of a trade union. The number of
women workers employed was 250, out of which 200 did not belong to any trade union.
In 2000, the number of union workers was 1725 of which 1600 were men. The number of non-union
workers was 380, among which 155 were women.
Solution.
TABLE 3·32. COMPARATIVE STUDY OF THE MEMBERSHIP OF
TRADE UNION IN A FACTORY IN 1995 AND 2000.
Year →
Trade Union
↓
Members
Non-members
Total
1995
Males
Females
1550 – 50
= 1,500
1,750 – 1500
= 250
2,000 – 250
= 1,750
250 – 200
= 50
2000
Total
Males
200
1,550
2000 – 1550
= 450
250
2,000
1,600
380 – 155
= 225
1,600 + 225
= 1,825
Females
Total
1,725 – 1600
= 125
1,725
155
125 + 155
= 280
380
1,725 + 380
= 2,105
Note. The bold figures are the given figures. The other values are obtained on appropriate additions or
subtractions, since the totals are fixed.
Example 3·13. In a sample study about coffee habit in two towns, the following information was
received :
Town A : Females were 40% ; Total coffee drinkers were 45% and Males non-coffee drinkers were 20%.
Town B : Males were 55% ; Males non-coffee drinkers were 30% and Females coffee drinkers were 15%.
Present the above data in a tabular form.
[C.A. (Foundation), May 1997]
3·33
CLASSIFICATION AND TABULATION
TABLE 3·33
Solution.
Town A
Coffee drinkers
Non-coffee drinkers
Males
60 – 20 = 40
20
Females
45 – 40 = 5
40 – 5 = 35
Total
45
100 – 45 = 55
Total
100 – 40 = 60
40
100
Note. The figures in bold are the given figures.
TABLE 3·33(a)
Town B
Coffee drinkers
Non-coffee drinkers
Males
Females
Total
55 – 30 = 25
30
15
60 – 30 = 30
25 + 15 = 40
100 – 40 = 60
55
100 – 55 = 45
100
Total
Note. The figures in the bold are the given figures.
The information in Tables 3·33 and 3·33(a) can be expressed in a single table as given in Table 3·33(b).
TABLE 3·33(b). SEX-WISE PERCENTAGE OF COFFEE DRINKERS IN TOWNS A AND B
Town A
Town B
Males
Females
Total
Males
Females
Total
Coffee drinkers
Non-coffee drinkers
40
20
5
35
45
55
25
30
15
30
40
60
Total
60
40
100
55
45
100
Example 3·14. Tabulate the following :
Out of a total number of 10,000 candidates who applied for jobs in a government department, 6,854
were males, 3,146 were graduates and others, non-graduates. The number of candidates with some
experience was 2,623 of whom 1,860 were males. The number of male graduates was 2,012. The number of
graduates with experience was 1,093 that includes 323 females.
Solution. We are given that the total number of :
Applicants = 10,000 ; Males = 6,854 ; Graduates = 3,146 ; Experienced = 2,623.
Total number of Females = 10,000 – 6,854 = 3,146
Total number of Non-graduates = 10,000 – 3,146 = 6,854
Total number of In-experienced persons = 10,000 - 2,623 = 7,377
The above and the remaining given information can be summarised in the following Table 3·34.
TABLE 3·34.DISTRIBUTION OF CANDIDATES FOR GOVERNMENT JOBS
SEX-WISE EDUCATION-WISE AND EXPERIENCE-WISE
Sex
↓
Male
Female
Total
Graduates
Non-graduates
Total
Experinced
In-experienced
Total
Experienced
In-experienced
Total
Experienced
In-experienced
Total
770
323
1242
811
2012
1134
1090
440
3752
1572
4842
2012
1860
763
4994
2383
6854
3146
1093
2053
3146
1530
5324
6854
2623
7377
10000
The figures in ‘bold’ are the given figures. The remaining values have been obtained by minor
calculations (additions or subtractions), as the totals are fixed.
Example 3·15. A survey of 370 students from Commerce Faculty and 130 students from Science
Faculty revealed that 180 students were studying for only C.A. Examinations, 140 for only Costing
Examinations and 80 for both C.A. and Costing Examinations.
3·34
BUSINESS STATISTICS
The rest had offered part-time Management Courses. Of those studying for Costing only, 13 were girls
and 90 boys belonged to Commerce Faculty. Out of 80 studying for both C.A. and Costing, 72 were from
Commerce Faculty amongst which 70 were boys. Amongst those who offered part-time Management
Courses, 50 boys were from Science Faculty and 30 boys and 10 girls from Commerce Faculty. In all there
were 110 boys in Science Faculty.
Present the above information in a tabular form. Find the number of students from Science Faculty
studying for part-time Management Courses.
Solution. We are given that :
Total number of Commerce students = 370
Total number of Science students = 130
∴ Total number of all the students = 370 + 130 = 500
We are also given that out of these 500 students, the number of students studying :
For C.A. only = 180 ; For Costing only = 140 ; For both Costing and C.A. = 80
∴ Number of students studying for part-time Management courses = 500 – (180 + 140 + 80) = 100.
The above information and the remaining given information is summarised in the Table 3·35. The
figures in ‘bold’ are the given figures.
TABLE 3·35. FACULTY, SEX AND COURSE-WISE DISTRIBUTION OF STUDENTS
Faculty
Courses
Part-time
Management
C.A. only
Costing only
C.A. and Costing
Total
Commerce
Science
Boys
Girls
30
10
Total
30 + 10
= 40
90
70
72 – 70 = 2
72
Boys
Girls
50
10
Total
Total
100 – 40
= 60
Boys
30 + 50
= 80
Girls
10 + 10
= 20
80 – 72 = 8
100
180
140
80
130
500
13
370
110
130 –
110 = 20
Total
Number of students from Science Faculty studying part-time Management courses is 60.
Example 3·16. In 1990, out of a total of 2,000 students in a college 1,400 were for Graduation and the
rest for Post-Graduation (P.G.). Out of 1,400 Graduate students 100 were girls. However, in all there were
600 girls in the college. In 1995, number of graduate students increased to 1,700, out of which 250 were
girls, but the number of P.G. students fell to 500 of which only 50 were boys. In 2000, out of 800 girls, 650
were for Graduation, whereas the total number of graduates was 2,200. The number of boys and girls in
P.G. classes was equal.
Represent the above information in tabular form. Also calculate the percentage increase in the number
of graduate students in 2000 as compared to 1990.
[C.A. (Foundation), Nov. 2001]
Solution. The distribution of the number of students with respect to level of education and sex is
obtained as follows :
Year 1990
Girls
Boys
Total
Graduation
100
1400 – 100 = 1300
1400
Post-Graduation
600 – 100 = 500
600 – 500 = 100
2000 – 1400 = 600
Total
600
2000 – 600 = 1400
2000
Note. The figures in bold are the given figures.
Year 1995
Girls
Boys
Total
Graduation
250
1,700 – 250 = 1450
1700
Post-Graduation
500 – 50 = 450
50
500
Total
250 + 450 = 700
1450 + 50 = 1500
1700 + 500 = 2200
3·35
CLASSIFICATION AND TABULATION
Year 2000
Graduation
650
2200 – 650 = 1550
2200
Girls
Boys
Total
Post-Graduation
800 – 650 = 150
150
150 + 150 = 300
Total
800
1550 + 150 = 1700
2500
The information in the above three tables can be expressed in single table as given in Table 3·36.
TABLE 3·36. DISTRIBUTION OF STUDENTS ACCORDING TO
DEGREE AND SEX FOR YEARS 1990 TO 2000.
Degrees →
Graduation
Post-graduation
Total
(a) + (b)
Boys
Girls
Total
(a)
Boys
Girls
Total
(b)
1990
1995
2000
1300
1450
1550
100
250
650
1400
1700
2200
100
50
150
500
450
150
600
500
300
2000
2200
2500
Total
4300
1000
5300
300
1100
1400
6700
Year ↓
Percentage increase in the number of graduate students in 2000 as compared to 1990 is :
(2200 – 1400)
1400
× 100 = 57·14%.
Example 3·17. Out of a total number of 1,807 women who were interviewed for employment in a
textile factory of Mumbai; 512 were from textile areas and the rest from the non-textile areas. Amongst the
married women who belonged to textile areas, 247 were experienced and 73 inexperienced, while for nontextile areas, the corresponding figures were 49 and 520. The total number of inexperienced women was
1,341 of whom 111 resided in textile areas. Of the total number of women, 918 were unmarried and of
these the number of experienced women in the textile and non-textile areas was 154 and 16 respectively.
Tabulate.
Solution. Total number of women interviewed = 1,807
No. of women from textile areas = 512
∴ Number of women from non-textile areas = 1,807 – 512 = 1,295
Total number of married women in textile areas = 247 + 73 = 320
Total number of married women in non-textile areas = 49 + 520 = 569
Total number of inexperienced women = 1,341
∴ Total number of experienced women = 1,807 – 1,341 = 466
Total number of unmarried women = 918
∴ Total number of married women = 1,807 – 918 = 889
Total number of unmarried experienced women in textile areas = 154
and Total number of unmarried experienced women in non-textile areas = 16
After filling this information in the table, the remaining entries in the table of the experience, marital
status and area-wise distribution of the number of women can now be completed by subtraction/addition,
wherever necessary and is given in Table 3·37.
TABLE 3·37. TABLE SHOWING THE NUMBER OF WOMEN INTERVIEWED FOR EMPLOYMENT
IN A TEXTILE FACTORY ACCORDING TO THEIR MARITAL STATUS,
EXPERIENCE AND AREA THEY BELONG
Textile Areas
Married
Unmarried
Total
Experienced
247
154
401
Inexperienced
73
38
111
Non-textile Areas
Total
320
192
512
Experienced
49
16
65
Inexperienced
520
710
1,230
Total
569
726
1,295
Total
Experienced
296
170
466
Inexperienced
593
748
1,341
Total
889
918
1,807
3·36
BUSINESS STATISTICS
Example 3·18. Draw up a blank table to show the number of candidates sex-wise, appearing in the
Pre-university, First Year, Second Year and Third Year examinations of a university in the faculties of Art,
Science and Commerce in the year 2002.
Solution.
TABLE 3·38. DISTRIBUTION OF CANDIDATES APPEARING IN THE UNIVERSITY
EXAMINATIONS w.r.t. FACULTY, SEX AND EXAMINATION IN 2002
Faculty →
Sex →
Arts
M
F
Examination ↓
Science
SubTotal
M
F
Commerce
SubTotal
M
F
Total
SubTotal
M
F
Row
Totals
Pre-university
First Year
Second Year
Third Year
Column Totals
Note. M indicates Male ; F indicates Female.
Example 3·19. Prepare a blank table to show the exports of three companies A, B, C to five countries
U.K, U.S.A., U.S.S.R., France and West Germany, in each of the years 1995 to 1999.
Solution.
TABLE 3·39. EXPORTS OF THE COMPANIES A, B AND C TO FIVE COUNTRIES
FROM 1995 TO 1999 (IN MILLION RUPEES)
Year →
Company →
Countries ↓
1995
A
B
1996
C
A
B
1997
C
A
B
1998
C
A
B
1999
C
A
B
Total
C
A
B
C
UK
USA
USSR
France
West Germany
Total
Example 3·20. Draw a blank table to present the following information regarding the college students
according to :
(a) Faculty : Social Sciences, Commercial Sciences.
(b) Class : Under-graduate and Post-graduate classes.
(c) Sex : Male and Female.
(d) Years : 1998 to 2002.
Solution. Please see Table 3·40, on page 3·37.
Example 3·21. Prepare a blank table showing the number of employees in a big business concern
according to :
(a) Sex : Males and Females.
(b) Five age-groups: Below 25 years, 25 to 35 years, 35 to 45 years, 45 to 55 years, 55 years and over.
(c) Two years: 2000 and 2001.
(d) Three grades of weekly salary: Below Rs. 4000; Rs. 4000 to 7000 ; Rs. 7000 and above.
Solution. Please see Table 3·41, on page 3·37.
TABLE 3·40. DISTRIBUTION OF COLLEGE STUDENTS ACCORDING TO SEX, CLASS AND FACULTY FOR 1998 TO 2002
Faculty →
Class →
Sex
→
Years
↓
Social Sciences
Commercial Sciences
Under-Graduate
Post-Graduate
Total
Under-Graduate
M F
S.T.
M F
S.T.
M
F
S.T. M
F
S.T.
(ii) + (iii) (vii) (viii) (ix )
(i) (ii)
(iii)
(iv) (v)
(vi)
(i)
= (i) + (ii)
= (iv) + (v) + (iv) (v) + (vi)
= (vii)
+ (viii)
M
(x)
Post-Graduate
F
S.T.
(xi)
(xii)
= (x) +
(xi)
M
(vii)
+ (x)
Total
F
S.T.
(ix)
(viii)
+ (xi) + (xii)
1998
1999
2000
2001
2002
Note : M indicates Males ; F Indicates Females ; S.T. indicates sub-totals.
Remark : In the caption, we may have another sub-group named ‘Total’ of Social Sciences and Commercial Sciences as in Example 3·22.
TABLE 3·41. DISTRIBUTION OF THE NUMBER OF EMPLOYEES OF A BUSINESS HOUSE
ACCORDING TO SEX, AGE AND SALARY FOR 2000 AND 2001.
Sex →
Likely salary
Years (’00 Rs.) →
↓
Age in years ↓
0—25
25—35
2000
35—45
45—55
55 and over
Sub-total
0—25
25—35
2001
35—45
45—55
55 and over
Sub-total
Male
0—40
40—70 70 and over Sub-total
Female
0—40
40—70
70 and over Sub-total
Total
0—40
40—70 70 and over Total
3·38
BUSINESS STATISTICS
EXERCISE 3·2.
1. (a) Explain the terms ‘classification’ and ‘tabulation’ and point out their importance in a statistical investigation.
What precautions would you take in tabulating statistical data ?
(b) What are the chief functions of tabulation ? What precautions would you take in tabulating statistical data ?
(c) Explain the purpose of classification and tabulation of data. State the rules that serve as a guide in tabulation of
data.
2. (a) What do you mean by tabulation of data ? What precautions would you take while tabulating data ?
(b) Distinguish between classification and tabulation of statistical data. Mention the requisites of a good statistical
table.
[Himachal Pradesh Univ. B.Com., 1998]
(c) Distinguish between classification and tabulation. What precautions would you take in tabulating data ?
3. (a) What do you understand by tabulation ? State any six points that should be kept in mind while tabulating the
data.
[CA (Foundation), May 1995]
(b) Which important points should be kept in mind while preparing a good statistical table ?
[C.A. (Foundation), May 1999]
(c) What are the rules to be followed in preparing a statistical table ?
4. (a) Briefly discuss the essential parts of a statistical table.
[C.A. (Foundation), May 2001]
(b) Draw a specimen table showing the various parts of a table.
[Osmania Univ. B.Com., 1997]
5. Comment on the statement : “In collection and tabulation of data common sense is the chief requisite and
experience the chief teacher”.
6. “The statistical table is a systematic arrangement of numerical data presented in columns and rows for purposes
of comparison.” Explain and discuss the various types of tables used in statistical investigation after the data have been
collected.
7. In a trip organised by a college there were 80 persons each of whom paid Rs. 15·50 on an average. There were
60 students each of whom paid Rs. 16. Members of the teaching staff were charged at a higher rate. The number of
servants was 6 (all males) and they were not charged anything. The number of ladies was 24% of the total of which one
was a lady staff member.
Tabulate the above information.
8. Tabulate the following data :
A survey was conducted amongst one lakh spectators visiting on a particular day cinema houses showing criminal,
social, historical, comic and mythological films. The proportion of male to female spectators under survey was three to
two. It indicated that while the respective percentages of spectators seeing criminal, social and historical films was
sixteen, twenty-six and eighteen, the actual number of female viewers seeing these types was four thousand six
hundred, twelve thousand two hundred, and seven thousand eight hundred respectively. The remaining two types of
films, namely, comic and mythological, were seen by forty per cent and one per cent of the male spectators. The
number of female spectators seeing mythological films was four thousand four hundred.
9. Present the following information in a suitable tabular form.
“In 1990, out of a total of 1750 workers of a factory, 1200 were members of a trade union. The number of women
employees was 200 of which 175 did not belong to a trade union. In 1995, the number of union workers increased to
1580 of which 1290 were men. On the other hand, the number of non-union workers fell to 208, of which 180 were
men.
In 2000, there were 1800 employees who belonged to a trade union and 50 who did not belong to a trade union. Of
all the employees in 2000, 300 were women, of whom only 8 did not belong to a trade union”.
10. Present the following information in a tabular form :
In 2001, out of a total of 4,000 workers in a factory, 3,300 were members of a trade union. The number of women
workers employed was 500 out of which 400 did not belong to the union. In 2000, the number of workers in the union
was 3,450 of which 3,200 were men. The number of non-union workers was 760 of which 330 were women.
11. A classification of the population of India by livelihood categories (agricultural and non-agricultural)
according to the 1951 census showed that out of total of 356,628 thousand persons, 249,075 thousand persons belonged
to agricultural category. In the agricultural category 71,049 thousand persons were self-supporting, 31,069 thousand
were earning dependents and the rest were non-earning. The number of non-earning persons and self-supporting
persons in the non-agricultural category were 67,335 thousand and 33,350 thousand respectively. The others were
earning dependents.
CLASSIFICATION AND TABULATION
3·39
Tabulate the above information expressing all figures in millions (1 million = 1,000 thousand).
12. A survey was conducted among 1,00,000 music listeners who were asked to indicate their preference for
classical music, light music, folk songs, film songs and pop varieties of music. The male listeners interviewed were as
many as female listeners. The survey indicated that while the percentage of listeners who preferred classical music,
light music and folk songs were eight, thirteen and four respectively; the actual number of females for each of the first
two kinds were six thousand. Of the listeners who liked folk songs, the number of male listeners was same as that of
female listeners. While film songs were liked by number one and half times that for all other varieties put together, the
number for pop music were only a fourth of the number of film song listeners. Sixty per cent of the listeners of pop
music were females.
Prepare a table showing the distribution of music listeners according to sex and type of music.
13. What are different parts of a table ? What points should be borne in mind while arranging the items in a table ?
An investigation conducted by the education department in a public library revealed the following facts. You are
required to tabulate the information as neatly and clearly as you can :
“In 1990, the total number of readers was 46,000 and they borrowed some 16,000 volumes. In 2000, the number of
books borrowed increased by 4,000 and the borrowers by 5%.
The classification was on the basis of three sections : literature, fiction and illustrated news. There were 10,000
and 30,000 readers in the sections literature and fiction respectively in the year 1990. In the same year 2,000 and 10,000
books were lent in the sections illustrated news and fiction respectively. Marked changes were seen in 2000. There
were 7,000 and 42,000 readers in the literature and fiction sections respectivley. So also 4,000 and 13,000 books were
lent in the sections illustrated news and fiction respectively.”
14. What is tabulation ? What are its uses ? Mention the items that a good statistical table should contain.
[C.A. (Foundation), Nov. 1996]
15. Out of total number of 2,807 women, who were interviewed for employment in a textile factory, 912 were
from textile areas and the rest from non-textile areas. Amongst the married women, who belonged to textile areas, 347
were having some work experience and 173 did not have work experience, while for non-textile areas the
corresponding figures were 199 and 670 respectively. The total number of women having no experience was 1,841 of
whom 311 resided in textile areas. Of the total number of women, 1,418 were unmarried and of these the number of
women having experience in the textile and non-textile areas was 254 and 166 respectively.
Tabulate the above information.
[C.A. (Foundation), May 1998]
16. In 1995 out of a total of 4,000 workers in a factory 3,300 were members of a trade union. The number of
women workers was 500 out of which 400 did not belong to the union. In 1994, the number of workers in the union
was 3,450 of which 3,200 were men. The number of workers not belonging to the union was 760 of which 300 were
women. Present data in a suitable tabular form.
[C.A. (Foundation), May 2000]
17. What are the considerations to be taken into account in the construction of a table ? Construct a table for
showing the profits of a company for a period of 5 years with imaginary figures.
[Madras Univ. B.Com., 1998]
18. (a) What are the components of a good table.
(b) Construct a blank table in which could be shown, at two different dates and in five industries, the average
wages of the four groups, males and females, eighteen years and over, and under eighteen years. Suggest a suitable
title.
19. State briefly the requirements of a good statistical table.
Prepare a blank table to show the distribution of population of various States and Union Territories of India
according to sex and literacy.
20. Draft a blank table to show the distribution of personnel working in an office according to (i) sex, (ii) three
grades of monthly salary - below Rs. 10,000 ; Rs. 10,000 to Rs. 20,000 ; above Rs. 20,000, (iii) age groups : below 25
years, 25—40 and 40—60 and (iv) 3 years : 1995-96, 1996-97, 1997-98.
21. Draw up a blank table to show five categories of skilled and unskilled workers i.e., regular, seasonal, casual,
clerical and supervisory; further divided into family members and paid workers with monthly/daily rate and piece rate.
22. Draw up in detail, with proper attention to spacing, double lines, etc., and showing all sub-totals, a blank table
in which could be entered the numbers occupied in six industries on two dates, distinguishing males from females, and
among the latter single, married and widowed.
3·40
BUSINESS STATISTICS
23. Draft a blank table to show the following information for the country A to cover the years 1974, 1989, 1999
and 2002.
(a) Population
;
(b) Income-tax collected
;
(c) Tobacco duties collected
(d) Spirits and beer duties collected
; (e) Other taxation.
Arrange for suitable columns to show also the “per capita” figures for (b), (c), (d), (e). Suggest a suitable title.
24. Fill in the blanks :
(i) … … is the first step in tabulation.
(ii) A … … is the systematic arrangement of data in rows and columns.
(iii) The numerical information in a statistical table is called the …… of the table.
(iv) In a statistical table, …… refer to the row headings and …… refer to column headings.
(v) In the collection and tabulation, …… is the chief requisite and …… is the chief teacher.
(vi) In a statistical table, the principal basis for the arrangement of captions and stubs in a systematic order are …
… , … … , and … …
(vii) Classification is the … … step in … …
(viii) In a statistical table, captions refer to the … … headings and stubs refer to the … … headings.
(ix) In a statistical table, the data are arranged in … … and … …
(x) In a statistical table, ……should be avoided, especially in titles and headings.
Ans. (i) classification ; (ii) table ; (iii) body ; (iv) stubs, captions ; (v) commonsense, experience.
(vi) alphabetical, chronological and geographical; (vii) first, tabulation ;
(viii) column, row ; (ix) rows and columns ; (x) abbreviations.
4
Diagrammatic and
Graphic Representation
4·1. INTRODUCTION
In Chapter 3, we discussed that classification and tabulation are the devices of presenting the statistical
data in neat, concise, systematic and readily comprehensible and intelligible form, thus highlighting the
salient features. Another important, convincing, appealing and easily understood method of presenting the
statistical data is the use of diagrams and graphs. They are nothing but geometrical figures like points, lines,
bars, squares, rectangles, circles, cubes, etc., pictures, maps or charts.
Diagrammatic and graphic presentation has a number of advantages, some of which are enumerated
below :
(i) Diagrams and graphs are visual aids which give a bird’s eye view of a given set of numerical data.
They present the data in simple, readily comprehensible form.
(ii) Diagrams are generally more attractive, fascinating and impressive than the set of numerical data.
They are more appealing to the eye and leave a much lasting impression on the mind as compared to the
dry and uninteresting statistical figures. Even a layman, who has no statistical background can understand
them easily.
(iii) They are more catching and as such are extensively used to present statistical figures and facts in
most of the exhibitions, trade or industrial fairs, public functions, statistical reports, etc. Human mind has a
natural craving and love for beautiful pictures and this psychology of the human mind is extensively
exploited by the modern advertising agencies who give their advertisements in the shape of attractive and
beautiful pictures. Accordingly diagrams and graphs have universal applicability.
(iv) They register a meaningful impression on the mind almost before we think. They also save lot of
time as very little effort is required to grasp them and draw meaningful inferences from them. An individual
may not like to go through a set of numerical figures but he may pause for a while to have a glance at the
diagrams or pictures. It is for this reason that diagrams, graphs and charts find a place almost daily in
financial/business columns of the newspapers, economic and business journals, annual reports of the
business houses, etc.
(v) When properly constructed, diagrams and graphs readily show information that might otherwise be
lost amid the details of numerical tabulations. They highlight the salient features of the collected data,
facilitate comparisons among two or more sets of data and enable us to study the relationship between them
more readily.
(vi) Graphs reveal the trends, if any present in the data more vividly than the tabulated numerical
figures and also exhibit the way in which the trends change. Although this information is inherent in a
table, it may be quite difficult and time-consuming (and sometimes may be impossible) to determine the
existence and nature of trends from a tabulation of data.
4·2. DIFFERENCE BETWEEN DIAGRAMS AND GRAPHS
No hard and fast rules exist to distinguish between diagrams and graphs but the following points of
difference may be observed :
(i) In the construction of a graph, generally graph paper is used which helps us to study the
mathematical relationship (though not necessarily functional) between the two variables. On the other hand,
4·2
BUSINESS STATISTICS
diagrams are generally constructed on a plane paper and are used for comparisons only and not for studying
the relationship between the variables. In diagrams data are presented by devices such as bars, rectangles,
squares, circles, cubes, etc., while in graphic mode of presentation points or lines of different kinds (dots,
dashes, dot-dash, etc.), are used to present the data.
(ii) Diagrams furnish only approximate information. They do not add anything to the meaning of the
data and, therefore, are not of much use to a statistician or research worker for further mathematical
treatment or statistical analysis. On the other hand, graphs are more obvious, precise and accurate than the
diagrams and are quite helpful to the statistician for the study of slopes, rates of change and estimation,
(interpolation and extrapolation), wherever possible. In fact, today, graphic work is almost a must in any
research work pertaining to the analysis of economic, business or social data.
(iii) Diagrams are useful in depicting categorical and geographical data but they fail to present data
relating to time series and frequency distributions. In fact, graphs are used for the study of time series and
frequency distributions.
(iv) Construction of graphs is easier as compared to the construction of diagrams.
In the following sections we shall first discuss the various types of diagrams and then the different
modes of graphic presentation.
4·3. DIAGRAMMATIC PRESENTATION
4·3·1. General Rules for Constructing Diagrams
1. Neatness. As already pointed out, diagrams are visual aids for presentation of statistical data and are
more appealing and fascinating to the eye and leave a lasting impression on the mind. It is, therefore,
imperative that they are made very neat, clean and attractive by proper size and lettering; and the use of
appropriate devices like different colours, different shades (light and dark), dots, dashes, dotted lines,
broken lines, dots and dash lines, etc., for filling the in between space of the bars, rectangles, circles, etc.,
and their components. Some of the commonly used devices are given below :
Fig. 4·1.
2. Title and Footnotes. As in the case of a good statistical table, each diagram should be given a
suitable title to indicate the subject-matter and the various facts depicted in the diagram. The title should be
brief, self explanatory, clear and non-ambiguous. However, brevity should not be attempted at the cost of
clarity. The title should be neatly displayed either at the top of the diagram or at its bottom.
If necessary the footnotes may be given at the left hand bottom of the diagram to explain certain points
or facts, not otherwise covered in the title.
3. Selection of Scale. One of the most important factors in the construction of diagrams is the choice of
an appropriate scale. The same set of numerical data if plotted on different scales may give the diagrams
differing widely in size and at times might lead to wrong and misleading interpretations. Hence, the scale
should be selected with great caution. Unfortunately, no hard and fast rules are laid down for the choice of
scale. As a guiding principle the scale should be selected consistent with the size of the paper and the size
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
4·3
of the observations to be displayed so that the diagram obtained is neither too small nor too big. The size of
the diagram should be reasonable so as to focus attention on the salient features and important
characteristics of the data. The scale showing the values should be in even numbers or multiples of 5 or 10.
The scale(s) used on both the horizontal and vertical axes should be clearly indicated. For comparative
study of two or more diagrams, the same scale should be adopted to draw valid conclusions.
4. Proportion Between Width and Height. A proper proportion between the dimensions (height and
width) of the diagram should be maintained, consistent with the space available. Here again no hard and
fast rules are laid down. In this regard Lutz in his book Graphic Presentation has suggested a rule called
‘root two’ rule, viz., the ratio 1 to √2 or 1 to 1·414 between the smaller side and the larger side respectively.
The diagram should be generally displayed in the middle (centre) of the page.
5. Choice of a Diagram. A large number of diagrams (discussed below) are used to present statistical
data. The choice of a particular diagram to present a given set of numerical data is not an easy one. It
primarily depends on the nature of the data, magnitude of the observations and the type of people for whom
the diagrams are meant and requires great amount of expertise, skill and intelligence. An inappropriate
choice of the diagram for the given set of data might give a distorted picture of the phenomenon under
study and might lead to wrong and fallacious interpretations and conclusions. Hence, the choice of a
diagram to present the given data should be made with utmost caution and care.
6. Source Note and Number. As in the case of tables, source note, wherever possible should be
appended at the bottom of the diagram. This is necessary as, to the learned audience of Statistics, the
reliability of the information varies from source to source. Each diagram should also be given a number for
ready reference and comparative study.
7. Index. A brief index explaining various types of shades, colours, lines and designs used in the
construction of the diagram should be given for clear understanding of the diagram.
8. Simplicity. Lastly, diagrams should be as simple as possible so that they are easily understood even
by a layman who does not have any mathematical or statistical background. If too much information is
presented in a single complex diagram it will be difficult to grasp and might even become confusing to the
mind. Hence, it is advisable to draw more simple diagrams than one or two complex diagrams.
4·3·2. Types of Diagrams. A large variety of diagrammatic devices are used in practice to present
statistical data. However, we shall discuss here only some of the most commonly used diagrams which may
be broadly classified as follows :
(1) One-dimensional diagrams viz., line diagrams and bar diagrams.
(2) Two-dimensional diagrams such as rectangles, squares, and circles or pie diagrams.
(3) Three-dimensional diagrams such as cubes, spheres, prisms, cylinders and blocks.
(4) Pictograms.
(5) Cartograms.
4·3·3. One-dimensional Diagrams
A. LINE DIAGRAM
This is the simplest of all the diagrams. It consists in drawing vertical lines, each vertical line being
equal to the frequency. The variate (x) values are presented on a suitable scale along the X-axis and the
corresponding frequencies are presented on a suitable scale along Y-axis. Line diagrams facilitate
comparisons though they are not attractive or appealing to the eye.
Remark. Even a time series data may be presented by a line diagram, by taking time factors along
X-axis and the variate values along Y-axis.
Example. 4·1. The following data shows the number of accidents sustained by 314 drivers of a public
utility company over a period of five years.
Number of accidents:
0
1
2
3
4
5
6
7
8
9
10
11
Number of drivers :
82
44
68
41
25
20
13
7
5
4
3
2
Represent the data by a line diagram.
4·4
BUSINESS STATISTICS
Solution.
NUMBER OF DRIVERS
82
80
LINE DIAGRAM
68
70
60
50
44
40
30
41
25
20
20
13
10
0
1
7
5
4
3
2
2 3 4 5 6 7 8 9 10 11
NUMBER OF ACCIDENTS
Fig. 4·2.
B. BAR DIAGRAM
Bar diagrams are one of the easiest and the most commonly used devices of presenting most of the
business and economic data. These are especially satisfactory for categorical data or series. They consist of
a group of equidistant rectangles, one for each group or category of the data in which the values or the
magnitudes are represented by the length or height of the rectangles, the width of the rectangles being
arbitrary and immaterial. These diagrams are called one-dimensional because in such diagrams only one
dimension viz., height (or length) of the rectangles is taken into account to present the given values. The
following points may be borne in mind to draw bar diagrams.
(i) All the bars drawn in a single study should be of uniform (though arbitrary) width depending on the
number of bars to be drawn and the space available.
(ii) Proper but uniform spacing should be given between different bars to make the diagram look more
attractive and elegant.
(iii) The height (length) of the rectangles or bars are taken proportional to magnitude of the
observations, the scale being selected keeping in view the magnitude of the largest observation.
(iv) All the bars should be constructed on the same base line.
(v) It is desirable to write the figures (magnitudes) represented by the bars at the top of the bars to
enable the reader to have a precise idea of the value without looking at the scale.
(vi) Bars may be drawn vertically or horizontally. However, in practice, vertical bars are generally used
because they give an attractive and appealing get up.
(vii) Wherever possible the bars should be arranged from left to right (from top to bottom in case of
horizontal bars) in order of magnitude to give a pleasing effect.
Types of Bar Diagrams. The following are the various types of bar diagrams in common use :
(a) Simple bar diagram.
(b) Sub-divided or component bar diagram.
(c) Percentage bar diagram.
(d) Multiple bar diagram.
(e) Deviation or Bilateral bar diagram.
(a) SIMPLE BAR DIAGRAM
Simple bar diagram is the simplest of the bar diagrams and is used frequently in practice for the
comparative study of two or more items or values of a single variable or a single classification or category
of data. For example, the data relating to sales, profits, production, population, etc., for different periods
4·5
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
may be presented by bar diagrams. As already pointed out the magnitudes of the observations are
represented by the heights of the rectangles.
Remark. If there are a large number of items or values of the variable under study, then instead of bar
diagram, line diagram may be drawn.
Example 4·2. The following data relating to the strength of the Indian Merchant Shipping Fleet gives
the Gross Registered Tonnage (GRT) as on 31st December, for different years.
Year :
1961
1966
1971
1975
1976
GRT in ’000 :
901
1,792
2,500
4,464
5,115
Source : Ministry of Shipping and Transport.
Represent the data by suitable bar diagram.
Solution.
STRENGTH OF INDIAN MERCHANT SHIPPING FLEET
GROSS REGISTERED TONNAGE IN ’000
6,000
5,115
4,464
4500
3000
2,500
1,792
1500
0
901
1961
1966
1971
1975 1976
Fig. 4·3.
(b) SUB-DIVIDED OR COMPONENT BAR DIAGRAM
A very serious limitation of the bar diagram is that it studies only one characteristic or classification at
a time. For example, the total number of students in a college for the last 5 years can be conveniently
expressed by simple bar diagrams but it cannot be used if we have also to depict the faculty-wise or sexwise distribution of students. In such a situation, sub-divided or component bar diagram is used. Subdivided bar-diagrams are useful not only for presenting several items of a variable or a category graphically
but also enable us to make comparative study of different parts or components among themselves and also
to study the relationship between each component and the whole.
In general sub-divided or component bar diagrams are to be used if the total magnitude of the given
variable is to be divided into various parts or sub-classes or components. First of all a bar representing the
total is drawn. Then it is divided into various segments, each segment representing a given component of
the total. Different shades or colours, crossing or dotting, or designs are used to distinguish the various
components and a key or index is given along with the diagram to explain these differences.
In addition to the general rules for constructing bar diagrams, the following points may be kept in mind
while constructing sub-divided or component bar diagrams :
(i) To facilitate comparisons the order of the various components in different bars should be same. It is
customary to show the largest component at the base of the bar and the smallest component at the top so
that the various components appear in the order of their magnitude.
(ii) As already pointed, an index or key showing the various components represented by different
shades, dottings, colours, etc., should be given.
4·6
BUSINESS STATISTICS
(iii) The use of sub-divided bar diagram is not suggested if the number of components exceeds 10,
because in that case the diagram is loaded with too much information and is not easy to understand and
interpret. Pie or circle diagram (discussed later) is appropriate in such a situation. The comparison of the
various components in different bars is quite tedious as they do not have a common base and requires great
skill and expertise.
Example 4·3. Represent the following data by a suitable diagram :
Items of Expenditure
Family A (Income Rs. 500)
Food
150
Clothing
125
Education
25
Miscellaneous
190
Saving or Deficit
+10
Family B (Income Rs. 300)
150
60
50
70
–30
Solution. The data can be represented by sub-divided bar diagram as shown below :
SUB-DIVIDED BAR DIAGRAM SHOWING
EXPENDITURE OF TWO FAMILIES
500
FOOD
400
300
CLOTHING
CLOTHING
FOOD
200
EDUCATION
CLOTHING
100
MISC.
EDUCATION
MISC.
0
–50
DEFICIT
SAVING
FAMILY A
FAMILY B
Fig. 4·4.
(c) PERCENTAGE BAR DIAGRAM
Sub-divided or component bar diagrams presented graphically on percentage basis give percentage bar
diagrams. They are specially useful for the diagrammatic portrayal of the relative changes in the data.
Percentage bar diagram is used to highlight the relative importance of the various component parts to the
whole. The total for each bar is taken as 100 and the value of each component or part is expressed, as
percentage of the respective totals. Thus, in a percentage bar diagram, all the bars will be of the same
height, viz., 100, while the various segments of the bar representing the different components will vary in
height depending on their percentage values to the total. Percentage bars are quite convenient and useful for
comparing two or more sets of data.
Item
Example 4·4. The adjoining table gives the break-up of
the expenditure of a family on different items of
consumption. Draw percentage bar diagram to represent the
data.
Food
Clothing
Rent
Fuel and Lighting
Education
Miscellaneous
Expenditure (Rs.)
240
66
125
57
42
190
4·7
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
Solution. First of all we convert the given figures into
percentages of the total expenditure as detailed below.
Rs.
Expenditure %
Cumulative %
240
240 × 100 =33·33
720
33·33
66
66
× 100 = 9·17
720
42·50
125
125 × 100 =17·36
720
59·86
Fuel and lighting
57
57 × 100 = 7·92
720
67·78
Education
42
42 × 100 = 5·83
720
73·61
Miscellaneous
190
190 × 100 =26·39
720
100·00
Total
720
100
Food
Clothing
Rent
Miscellaneous
26·39%
80
PERCENTAGE EXPENDITURE
Item
DIAGRAM SHOWING EXPENDITURE
OF FAMILY ON DIFFERENT ITEMS
OF CONSUMPTION
100
Education 5·83%
Fuel and Lighting
7·92%
60
Rent
17·36%
40
Clothing
9·17%
20
Food
33·33%
The percentage bar diagram is given in Fig. 4·5.
0
FAMILY
Fig. 4·5.
Example 4·5. Draw a bar chart for the following data showing the percentage of total population in
villages and towns:
Percentage of total population in
Villages
Towns
Infants and young children
13·7
12·9
Boys and girls
25·1
23·2
Young men and women
32·3
36·5
Middle-aged men and women
20·4
20·1
8·5
7·3
Elderly persons
Solution.
CALCULATIONS FOR PERCENTAGE BAR DIAGRAMS
Category
Villages
%
Towns
Cumulative %
%
Cumulative %
Infants and young children
13·7
13·7
12·9
12·9
Boys and girls
25·1
38·8
23·2
36·1
Young men and women
32·3
71·1
36·5
72·6
Middle aged men and women
20·4
91·5
20·1
92·7
8·5
100·0
7·3
100·0
Elderly persons
100
100
4·8
BUSINESS STATISTICS
PERCENTAGE POPULATION
PERCENTAGE BAR DIAGRAM SHOWING TOTAL POPULATION
IN VILLAGES AND TOWNS BY DIFFERENT CATEGORIES
100
90
80
Infants and Young Children
Boys and Girls
Young Men and Women
Middle Aged Men and Women
Elderly Persons
70
60
50
40
30
20
10
VILLAGES TOWNS
Fig. 4·6.
(d) MULTIPLE BAR DIAGRAM
A limitation of the simple bar diagram was that it can be used to portray only a single characteristic or
category of the data. If two or more sets of inter-related phenomena or variables are to be presented
graphically, multiple bar diagrams are used. The technique of drawing multiple bar diagram is basically
same as that of drawing simple bar diagram. In this case, a set of adjacent bars (one for each variable) is
drawn. Proper and equal spacing is given between different sets of the bars. To distinguish between the
different bars in a set, different colours, shades, dottings or crossings may be used and key or index to this
effect may be given.
Example 4·6. The data below give the yearly profits (in thousand of rupees) of two companies A and B.
Profits in (’000 rupees)
Year
1994-95
1995-96
1996-97
1997-98
1998-99
Company A
120
135
140
160
175
Company B
90
95
108
120
130
Represent the data by means of a suitable diagram.
Solution. The data can be suitably represented by a multiple bar diagram as shown below.
YEARLY PROFITS IN ’000 RUPEES
160
120
108
135
95
120
90
100
140
Company B
150
175
Company A
130
200
50
0
1994-95
1995-96
1996-97
Fig. 4·7.
1997-98
1998-99
4·9
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
Remark. A careful examination of the above figures of profits for the two companies A and B reveals
that in all the years from 1994-95 to 1998-99, company A shows higher profits than the company B. In such
a situation when the values of one concern or unit show an increase over the values of the other concern or
unit for all the periods under
YEARLY PROFITS IN ’000 RUPEES
200
consideration, the data can be
175
elegantly represented by a
160
special type of sub-divided 150
45
Company B
140
135
bar diagram, in which total
40
120
refers to the values of the
32
40
Excess
concern or unit with higher 100
30
values and the lower portion
(shaded) of the bar shows the
values of other concern. The
50
remaining portion (blank)
shows the balance (excess) of
0
the two concerns or units. We
1994-95 1995-96 1996-97 1997-98 1998-99
represent in Fig. 4.8 the
above data in this manner.
Fig. 4·8.
Example 4·7. The following data shows the students in millions on rolls at school/university stage in
India according to different class groups and sex for the year 1970-71 as on 31st March.
Stage
Boys
Girls
Total
35·74
21·31
57·05
Class VI to VIII
9·43
3·89
13·32
Class IX to XI
4·87
1·71
6·58
University/College
2·17
0·64
2·81
Class I to V
Represent the data by (i) Component bar diagram and (ii) Multiple bar diagram.
Solution.
Fig. 4·9.
35·74
Boys
5
Class
I-V
Class
VI-VIII
0·64
15
10
2·17
Girls
4·37
1·71
45
40
35
30
25
20
9·43
3·83
6·58
Class Class Class University/
I-V VI-VIII IX-XI College
50
21·31
Boys
13·32
20
15
10
5
Girls
(ii) MULTIPLE BAR DIAGRAM
SHOWING STUDENTS ON ROLL
AT SCHOOL/UNIVERSITY STAGE
ACCORDING TO SEX IN 1970-71
NUMBER OF STUDENTS IN MILLIONS
55
50
45
40
35
30
25
2·81
60
57·05
NUMBER OF STUDENTS IN MILLIONS
(i) COMPONENT BAR DIAGRAM
SHOWING STUDENTS ON ROLL
AT SCHOOL/UNIVERSITY STAGE
ACCORDING TO SEX IN 1970-71
Class University/
IX-XI
College
Fig. 4·10.
4·10
BUSINESS STATISTICS
(e) DEVIATION BARS
Deviation bars are specially useful for graphic presentation of net quantities viz., surplus or deficit, e.g.,
net profit or loss, net of imports and exports which have both positive and negative values. The positive
deviations (e.g., profits, surplus) are presented by bars above the base line while negative deviations (loss,
deficit) are represented by bars below the base line. The following example will illustrate the points.
Remark. Deviation bars are also sometimes known as Bilateral Bar Diagrams and are used to depict
plus (surplus) and minus (deficit) directions from the point of reference.
Example 4·8. For the following data prepare a suitable diagram showing Balance of Trade :
Years
1994
1995
1996
1997
1998
1999
Exports (In Rs. Million)
24
115
84
110
130
162
Inports (In Rs. Million)
9
92
92
120
183
187
[Delhi Univ. B.Com. (Pass), 2001]
Solution.
Year
1994
1995
1996
1997
1998
1999
Exports (In Rs. Million)
24
115
84
110
130
162
Imports (In Rs. Million)
9
92
92
120
183
187
Balance of Trade (In Million Rs.)
24 – 9 = 15
115 – 92 = 23
84 – 92 = – 8
110 – 120 = –10
130 – 183 = –53
162 – 187 = –25
Deviation bar diagram showing balance of trade is given in Fig. 4·11 :
Y
DEVIATION BAR DIAGRAM
BALANCE OF TRADE
(In Million Rs.)
30 –
20 –
23
15
10 –
0–
–10 –
1996
1994
1997
1998
1999
1995
–8
–10
–20 –
–25
–30 –
–40 –
–50 –
–53
–60 –
Fig. 4·11.
X
4·11
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
For more illustrations, see Examples 4·13 and 4·14
(f) BROKEN BARS
Broken bars are used for graphic presentation of the data which contain very wide variations in the
values i.e., the data which contain very large observations along with small observations. In this case the
squeezing of the vertical scale will not be of much help because it will make the small bars to look too
small and clumsy and thus will not reveal the true characteristics of the data. In order to provide adequate
and reasonable shape for smaller bars, the larger (or largest) bar(s) may be broken at the top, as illustrated
in Examples 4·9 and 4·10.
Remark. However, if all the observations are fairly large so that all the bars have a broken vertical
axis, then instead they can be drawn with a false base line for the vertical axis. [For false base line, see
§ 4·4·2—Graphic Presentation.]
Example 4·9. The following data relates to the imports of foreign merchandise and exports (including
re-exports) of Indian merchandise (in million rupees) for some countries for the year 1975-76.
Country
Imports
Burma
Exports
Country
Imports
Exports
53
89
Germany (F.R.)
3,566
1,173
522
343
Iran
4,593
2,708
Canada
2,278
424
United Kingdom
2,683
4,020
Australia
1,015
477
USSR
2,958
4,128
799
785
Japan
1,852
835
USA
Czechoslovakia
Italy
France
3,548
4,263
12,699
5,054
Represent the data by suitable diagram.
Solution. The above data is represented by bar-diagrams as shown below.
FOREIGN TRADE BY COUNTRIES
1975-76
IMPORTS
EXPORTS
(Including Re-Exports)
Burma
Czechoslovakia
Canada
Australia
Italy
France
Germany (F.R.)
Iran
United Kingdom
U.S.S.R.
Japan
1,270
U.S.A.
600
400
200
0
0
200
400
600
RUPEES TEN MILLION
RUPEES TEN MILLION
Fig. 4·12.
4·12
BUSINESS STATISTICS
Example 4·10. Represent the following data relating to the military statistics at the border during the
war between the two countries A and B in 1999 by multiple bar diagram.
Category
Country A
Army Divisions
Semi-Army Units
Fighter Planes
Tanks
Total Troops
Country B
4
50
75
50
100,000
20
—
700
300
170,000
Solution.
100,000
170,000
MILITARY STATISTICS AT THE BORDER
Country A
Country B
700
300
75
50
50
20
4
Army Semi-Army Fighter
Divisions Units
Planes
Tanks
Total
Troops
Fig. 4·13.
4·3·4. Two-dimensional Diagrams. Line or bar diagrams discussed so far are one-dimensional
diagrams since the magnitudes of the observations are represented by only one of the dimensions viz.,
height (length) of the bars while the width of the bars is arbitrary and uniform. However, in twodimensional diagrams, the magnitudes of the given observations are represented by the area of the diagram.
Thus, in the case of two-dimensional bar diagrams, the length as well as width of the bars will have to be
considered. Two-dimensional diagrams are also known as area diagrams or surface diagrams. Some of the
commonly used two-dimensional diagrams are :
(A)
(B)
(C)
(D)
Rectangles.
Squares.
Circles.
Angular or pie diagrams.
(A) Rectangles. A “rectangle” is a two-dimensional diagram because it is based on the area principle.
Since the area of a rectangle is given by the product of its length and breadth, in a rectangle diagram both
the dimensions viz., length (height) and width of the bars is taken into consideration.
Just like bars, the rectangles are placed side by side, proper and equal spacing being given between
different rectangles. In fact, rectangle diagrams are a modified form of bar diagrams and give a more
detailed information than bar diagrams.
4·13
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
COST AND PROFIT PER UNIT (IN RS.)
Like sub-divided bars, we have also sub-divided rectangles for depicting the total and its break-up into
various components. Likewise percentage rectangle diagram may be used to portray the relative
magnitudes of two or more sets of data and their components making up the total. We give below a few
illustrations.
DIAGRAM SHOWING COST AND PROFIT
Example 4·11. Prepare a rectangular diagram
FOR A COMMODITY IN A FACTORY
from the following particulars relating to the
9
production of a commodity in a factory.
Units produced
1,000
8
Cost of raw materials
Rs. 5,000
Raw
7
Material
Direct expenses
Rs. 2,000
6
Indirect expenses
Rs. 1,000
Profit
Rs. 1,000
5
Solution. First of all we will find the cost of
4
material, expenses and profits per unit as given
Direct
below :
3
5000
Cost of raw material per unit = Rs. 1000
= Rs. 5
Rs. 2000
1000 = Rs. 2
1000
Rs. 1000
= Re. 1
Direct expenses per unit
=
Indirect expenses per unit
=
1000
Profit per unit = Rs. 1000
= Re. 1
Expenses
2
Indirect
Expenses
1
0
Profit
1000
UNITS PRODUCED
Fig. 4·14.
Example 4·12. The following data relates to the monthly expenditure (in Rs.) of two families A and B.
Expenditure (in Rs.)
Item of Expenditure
Family A
160
80
60
20
80
400
Food
Clothing
Rent
Light and fuel
Miscellaneous
Total
Family B
120
32
48
16
24
240
Represent it by a suitable percentage diagram.
Solution. Since the total expenses of the two families are different, an appropriate percentage diagram
for the above data will be rectangular diagram on percentage basis. The percentage bar diagram will not be
able to reflect the inherent differences in the total expenditures of the two families.
The widths of the rectangles will be taken in the ratio of the total expenses of the two families viz.,
400 : 240 i.e., 5 : 3.
CALCULATIONS FOR PERCENTAGE RECTANGULAR DIAGRAM
Item of Expenditure
Food
Clothing
Rent
Light and fuel
Miscellaneous
Total
Rs.
160
80
60
20
80
400
Family A
%
40
20
15
5
20
100
Cumulative %
40
60
75
80
100
Rs.
120
32
48
16
24
240
Family B
%
50
13·33
20
6·67
10
100
Cumulative %
50
63·33
83·33
90
100
4·14
BUSINESS STATISTICS
PERCENTAGE RECTANGLE DIAGRAM SHOWING
MONTHLY EXPENDITURE OF TWO FAMILIES A AND B
100
90
80
Food
Food
70
60
50
Clothing
Clothing
40
30
20
10
Rent
Rent
Light and Fuel
Misc.
Light & Fuel
Misc.
0
Family A
Family B
Fig. 4·15.
Example 4·13. Represent the following data by a percentage sub-divided bar diagram.
Item of Expenditure
Family A
Income Rs. 500
150
125
25
190
+10
Food
Clothes
Education
Miscellaneous
Savings or Deficit
Family B
Income Rs. 300
150
60
50
70
–30
Solution. Since the total incomes of the two families are different, an appropriate percentage bar
diagram for the above data will be rectangular diagram on percentage basis. The percentage bar diagram
will not be able to reflect the inherent differences in the total incomes in the two families.
The widths of the rectangles will be taken in the ratio of the total incomes of the families viz., 500 : 300
i.e., 5 : 3.
CALCULATIONS FOR PERCENTAGE RECTANGULAR DIAGRAM
Item of Expenditure
Food
Clothes
Education
Miscellaneous
Savings or Deficit
Total
Expenditure
(Rs.)
150
125
25
190
+ 10
500
Family A
%
30
25
5
38
2
Cumulative %
30
55
60
98
10
Expenditure
(Rs.)
150
60
50
70
– 30
300
Family B
%
Cumulative %
50
20
16.7
23.3
– 10
50
70
86.7
110.0
100
4·15
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
PERCENTAGE EXPENDITURE
PERCENTAGE DIAGRAM SHOWING MONTHLY INCOME
AND EXPENDITURE OF TWO FAMILIES A AND B
Family B
Family A
100
Food
90
Clothes
30
80
Education
50
70
Misc.
60
25
50
5
20
38
16·7
40
30
20
10
23·3
0
2
Saving
Rs. 500
Rs. 300
Income ⎯
⎯→
10 Deficit
Fig. 4·16.
Example 4·14. Draw a suitable diagram to represent the following information.
Factory X
Factory Y
Selling price
per unit
(in Rs.)
Quantity
sold
400
600
20
30
Total Cost (in Rs.)
—————————————————————————————————————————————————————————————————————————————————
Wages
Materials
Misc.
Total
3,200
6,000
2,400
6,000
1,600
9,000
7,200
21,000
Show also the profit or loss as the case may be.
Solution. First of all we shall calculate the cost (wages, materials, misc.) and profit per unit as given in
the following table.
Selling price per
unit (in Rs.)
Quantity
sold
Cost per unit (in Rs.)
Wages
Materials
Misc.
Total
Profit per
unit (in Rs.)
Factory X
400
20
160
120
80
360
400 – 360 = 40
Factory Y
600
30
200
200
300
700
600 – 700 = 100
Note. Negative profit is regarded as loss.
An appropriate diagram for representing this data would be the ‘Rectangles’ whose widths are in the
ratio of the quantities sold i.e., 20 : 30 i.e., 2 : 3. Selling prices would be represented by the corresponding
heights of the rectangles with various factors of cost (wages, materials, misc.) and profit or loss represented
by the various divisions of the rectangles as shown in the following diagram (Fig. 4.17).
Remark. In the case of profit i.e., when selling price (S.P.) is greater than cost price (C.P.), the entire
rectangle will lie above the X-axis, the segment just above the X-axis showing profit. But in case of loss
i.e., when S.P. is less than C.P., we will have the rectangle with a portion lying below the X-axis which will
reflect the loss incurred i.e., the cost not recovered through sales. The values of each component are given
by the product of the base with the corresponding height of the component (rectangle). For example, for the
factory X, the area of the component for wages is 20 × 160 = 3200, which is the given cost.
4·16
BUSINESS STATISTICS
SUB-DIVIDED RECTANGLE SHOWING COST, SALES AND PROFIT OR LOSS PER UNIT
SALES (IN RUPEES)
800
Factory Y
30 Units
700
600
Factory X
20 Units
Wages
300
Wages
Materials
200
Materials
100
Misc.
Profits
500
400
0
Misc.
Loss
–100
Fig. 4·17.
(B) Square Diagrams. Among the two-dimensional diagrams, squares are specially useful if it is
desired to compare graphically the values or quantities which differ widely from one another as e.g., the
population of different countries at a given time or of the same country at different times or the imports
or exports of different countries. In such a situation, the bar diagrams are not suitable, since they will give
very disproportionate bars i.e., the bars corresponding to smaller quantities would be comparatively too
small and those corresponding to bigger values would be too big. In particular, if two values are in the
proportion of 1 : 25 and if we draw bar diagram, one (bigger) will be 25 times (in height) than that of the
other (smaller). In such a situation, square diagrams give a better presentation.
Like rectangle diagram, square diagram is a two-dimensional diagram in which the given values are
represented by the area of the square. Since the area of the square is given by the square of its side, the side
of the square diagram will be in proportion to the square root of the given observations. Thus if the two
observations are in the ratio of 1 : 25, the sides of the squares will be in the ratio of their square roots viz.,
1 : 5.
Construction of the square diagrams is quite simple. First of all we obtain the square roots of the given
observations and then squares are drawn with sides proportional to these square roots, on an appropriate
scale which must be specified.
Remarks 1. The square may be drawn horizontally (on the same base line) or vertically one below the
other to facilitate comparisons. However, in practice, the first method viz., horizontal presentation is
generally used since it economises space.
2. Although square diagram is a two-dimensional diagram, it is used to depict only a single magnitude
or value.
Example 4·15. Draw a square diagram to represent the following data.
Country
A
B
C
Yield in (kg.) per hectare
350
647
1,120
YIELD in kg. (Per hectare)
Solution. The square roots of the given yields
in (kg) per hectare give the proportion of the sides
SCALE :
of the corresponding squares. The calculations are
1 Square cm = 350 kg.
shown in the following table :
Country
Yield in (kg.)
per hectare
Square root
Ratio of the sides
of the squares
A
B
1·79 cm
C
1·36 cm
350
18·7083
647
25·4362
1
1·36
1,120
33·4664
1·79
The square diagram is given in Fig. 4·18.
1 cm
A
B
Fig. 4·18.
C
4·17
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
Remark. In the above table, the ratio of the sides of the squares has been obtained on dividing the
square roots of the values for B and C by the square root of the value for A. This is easy to do on a
calculator but without a calculator it is quite time-consuming. However, the things can be simplified to a
great extent by dividing the square roots of the values of A, B and C by a whole number, say, 15 or 18.
Division by, say, 15, gives the ratio of the sides of squares for A, B and C as 1·25, 1·70, 2·23 respectively.
Squares can now be constructed by taking appropriate scale which must be specified in the diagram.
If we construct squares with sides 1·25 cms., 1·70 cms., and 2·23 cms., the scale will be obtained as
follows :
Area of square for A is (1·25)2 = 1·5625 square cms. This area represents the value 350 kg. Thus,
1·5625 sq. cms. = 350 kg.
350
1 sq. cm. = 1·5625
= 224 kg.
⇒
(C) Circle Diagrams. Circle diagrams are alternative to square diagrams and are used for the same
purpose, viz., for diagrammatic presentation of the values differing widely in their magnitude. The area of
the circle, which represents the given values is given by πr2 , where π = 22/7 and r is the radius of circle. In
other words, the area of the circle is proportional to the square of its radius and consequently, in the
construction of the circle diagram the radius of the circle is a value proportional to the square root of the
given magnitude. Accordingly, the lengths which were taken as the sides of the square may also be taken as
the radii of the circles representing the given magnitudes.
Remarks 1. Circle diagrams are more attractive and appealing than square diagrams and since both
require more or less the same amount of work, viz., computing the square roots of the given magnitudes
(rather circles are easy to draw), circle diagrams are generally preferred to square diagrams.
2. Since square and circle diagrams are to be compared on an area basis, it is difficult to judge the
relative magnitudes with precision, particularly by a layman without any mathematical or statistical
background. Accordingly, proper care should be taken to interpret them. They are also more difficult to
construct than the rectangle diagrams.
3. Scale. The scale to be used for constructing circle diagrams can be calculated as follows :
For a given magnitude ‘a’ we have
a
Area = πr2 square units = a
⇒
1 square unit = 2 .
πr
Example 4·16. Represent the data of Example 4·15 by a circular diagram.
Solution. The data of Example 4·15 can be represented by a circular diagram on taking the lengths of
the sides of the squares which were taken in Example 4·15, as the radii of the corresponding circles.
However, in this case, the scale will be modified accordingly.
2450
Scale : 1 sq. cm. = 350
π = 22 = 111·36 kg.
YIELD IN KG (PER HECTARE)
SCALE : 1 sq. cm. = 111·36 kg.
1 cm
A
1·79 cm
1.36 cm
C
B
Fig. 4·19.
4·18
BUSINESS STATISTICS
(D) Angular or Pie Diagram. Just as sub-divided and percentage bars or rectangles are used to
represent the total magnitude and its various components, the circle (representing the total) may be divided
into various sections or segments viz., sectors representing certain proportion or percentage of the various
component parts to the total. Such a sub-divided circle diagram is known as an angular or pie diagram,
named so because the various segments resemble slices cut from a pie.
Steps for Construction of Pie Diagram
1. Express each of the component values as a percentage of the respective total.
2. Since the angle at the centre of the circle is 360°, the total magnitude of the various components is
taken to be equal to 360° and each component part is to be expressed proportionately in degrees. Since
1 per cent of the total value is equal to 360/100 = 3·6°, the percentage of the component parts obtained in
step 1 can be converted to degrees by multiplying each of them by 3·6.
3. Draw a circle of appropriate radius using an appropriate scale depending on the space available. If
only one category or characteristic is to be used, the circle may be drawn of any radius. However, if two or
more sets of data are to be presented simultaneously for comparative studies, then the radii of the
corresponding circles are to be proportional to the square roots of their total magnitudes.
4. Having drawn the circle, draw any radius (preferably horizontal). Now with this radius as the base
line draw an angle at the centre [with the help of protractor (D)] equal to the degree represented by the first
component, the new line drawn at the centre to form this angle will touch the circumference. The sector so
obtained will represent the proportion of the first component. From this second line as base, now draw
another angle at the centre equal to the degree represented by the 2nd component, to give the sector
representing the proportion of the second component. Proceeding similarly, all the sectors representing
different component parts can be constructed.
5. Different sectors representing various component parts should be distinguished from one another by
using different shades, dottings, colours, etc., or giving them explanatory or descriptive labels either inside
the sector (if possible) or just outside the circle with proper identification.
Remarks 1. The degrees represented by the various component parts of a given magnitude can be
obtained directly without computing their percentage to the total value as follows :
Degree of any component part = Component value × 360°.
Total value
2. Pie diagrams are also called circular diagrams.
3. Since the comparison of the pie diagrams is to be made on the basis of the areas of the circles and of
various sectors which are difficult to be ascertained visually with precision, generally sub-divided or
percentage bars or rectangles are preferred to pie diagrams for studying the changes in the total and
component parts. Moreover, pie diagrams are difficult to construct as compared with bars or rectangle
diagrams. Any way, if the number of component parts is more than 10, pie chart is preferred to bar or
rectangle diagram which becomes rather confusing in such a situation.
We give below some illustrations of pie diagrams.
Example 4·17. Draw a pie diagram to represent the following data of proposed expenditure by a State
Government for the year 1997-98.
Items
Proposed Expenditure
(in million Rs.)
Agriculture & Rural
Development
Industries & Urban
Development
Health &
Education
Miscellaneous
4,200
1,500
1,000
500
[Delhi Univ. B.Com. (Pass), 1997]
4·19
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
Solution.
CALCULATIONS FOR PIE CHART
Items
Proposed expenditure (in million Rs.)
(1)
(2)
Angle at the centre
(3) =
(2)
× 360°
7200
Agriculture and Rural Development
4,200
42
72 × 360° = 210°
Industries and Urban Development
1,500
15
72 × 360° = 75°
Health and Education
1,000
10
72 × 360° = 50°
500
5
72 × 360° = 25°
7,200
360°
Miscellaneous
Total
PIE DIAGRAM REPRESENTING PROPOSED EXPENDITURE
BY STATE GOVERNMENT ON DIFFERENT ITEMS FOR 1997-98
Agriculture and
Rural Development
Miscellaneous
Industries
and Urban
Development
Health and Education
Fig. 4·20.
Example 4·18. The following data shows the expenditure on various heads in the first three five-year
plans (in crores of rupees).
Expenditure (in crores Rs.)
Subject
First Plan
Second Plan
Third Plan
Agriculture and C.D.
361
529
1068
Irrigation and Power
561
865
1662
Village and Small Industries
173
176
264
Industry and Minerals
292
900
1520
Transport and Communications
497
1300
1486
Social Services and Miscellaneous
477
830
1500
2361
4600
7500
Total
Represent the data by angular (pie) diagrams.
4·20
BUSINESS STATISTICS
Solution.
CALCULATIONS FOR PIE DIAGRAMS
Expenditure (in crores Rupees)
Items of Expenditure
First Plan
Second Plan
Rs.
Degrees
361
361
2361 × 360° = 55·1°
529
529
4600 × 360° = 41·4°
1,068 7500 × 360° = 51·2°
561
561
2361 × 360° = 85·5°
865
865
4600 × 360° = 67.7°
1,662 7500 × 360° = 79·8°
173
173
2361 × 360° = 26·4°
176
176
4600 × 360° = 13·8°
292
292
2361 × 360° = 44·5°
900
900
4600 × 360° = 70·4°
1,520 7500 × 360° = 73·0°
497
497
2361 × 360° = 75·8°
1300
1300
4600 × 360° = 101·7°
1,486 7500 × 360° = 71·3°
477
477
2361 × 360° = 72·7°
830
830
4600 × 360° = 65·0°
1,500 7500 × 360° = 72·0°
Total
2,361
360°
4,600
360°
Square Root
48·59
67·82
86·60
1·0
1·4
1·8
Agriculture
and C.D.
Irrigation
and Power
Village and
Small Industries
Industry
and Minerals
Transport
and Communications
Social Services
and Miscellaneous
Radii of circles
Rs.
Third Plan
Degrees
Rs.
Degrees
1068
1662
264
264
7500 × 360° = 12·7°
1520
1486
1500
7,500
360°
THIRD PLAN
FIRST PLAN
SECOND PLAN
EXPENDITURE ON VARIOUS HEADS IN FIRST THREE FIVE-YEAR PLANS
Agriculture and C.D.
Industry and Minerals
Irrigation and Power
Transport and Communications
Village and Small Industries
Social Services and Miscellaneous
Fig. 4·21.
4·3·5. Three-Dimensional Diagrams. Three-dimensional diagrams, also termed as volume diagrams
are those in which three dimensions, viz., length, breadth and height are taken into account. They are
constructed so that the given magnitudes are represented by the volumes of the corresponding diagrams.
The common forms of such diagrams are cubes, spheres, cylinders, blocks, etc. These diagrams are
specially useful if there are very wide variations between the smallest and the largest magnitudes to be
represented. Of the various three-dimensional diagrams, ‘cubes’ are the simplest and most commonly used
devices of diagrammatic presentation of the data.
4·21
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
Cubes. For instance, if the smallest and the largest magnitudes to be presented are in the ratio of
1 : 1000, the bar diagrams cannot be used because the height of the biggest bar would be 1000 times the
height of the smallest bar and thus they would look very disproportionate and clumsy. On the other hand, if
square or circle diagrams are used then the sides (radii) of the squares (circles) will be in the ratio of the
square roots viz., 1 : √
⎯⎯⎯
1000 i.e., 1 : 31·63 i.e., 1 : 32 (approx.), which will again give quite disproportionate
diagrams. However, if cubes are used to present this data, then since the volume of cube of side x is x 3 , the
3
sides of the cubes will be in the ratio of their cube roots viz., 1 : √
⎯⎯⎯⎯
1000 i.e., 1 : 10, which will give
reasonably proportionate diagrams as compared to one-dimensional or two-dimensional diagrams.
Construction of a Cube of Side ‘x’. The various steps are outlined below :
E
1. Construct a square ABCD of side x.
2. Draw EF as right bisector i.e., perpendicular bisector
of AB, [This is done by finding the mid-point of AB and
then drawing perpendicular at that point to the line AB] such
that EF = AB and half of it is above AB and half of it is
below AB.
G
A
B
(iii) Join AE, CF and EF.
(iv) Through B draw a line BG parallel to AE (i.e.,
BG || AE) such that BG = AE.
(v) Join EG and through G draw a line GH || EF such
that GH = EF.
H
F
C
D
x
(vi) Join D and H.
Fig. 4·22
(vii) Rub off the lines CF, EF and FH. Now CDHGEAC
is the required cube.
Remarks 1. As already discussed, three-dimensional diagrams are used with advantage over one or
two-dimensional diagrams if the range i.e., the gap between the smallest and the largest observation to be
presented, is very large. Moreover, they are more beautiful and appealing to the eye than bars, rectangles,
squares or circles. However, since three-dimensional diagrams are quite difficult to construct and
comprehend as compared to one-or two-dimensional diagrams, they are not very popular. Further, as the
magnitudes are represented by the volumes of the cubes (volumes of the three-dimensional diagrams, in
general), it is very difficult to visualise and hence interpret them with precision.
2. Cylinders, spheres and blocks are quite difficult to construct and are, therefore, not discussed here.
3. It is worthwhile pointing out here that now-a-days projection techniques are used to represent
even one-dimensional diagrams as three-dimensional diagrams for giving them a beautiful and attractive
get up.
Example 4·19. The following table gives the population of India on the basis of religion.
Religion
Number (in lakhs)
Hinduism
2031·9
Islam
354·0
Christianity
81·6
Sikhism
62·2
Others
36·3
Represent the data by cubes.
Solution. The sides of the cubes will be proportional to the cube roots of the magnitudes they
represent.
Note. To compute cube root of a, viz., √
⎯3 a or a1/3,
Taking logarithm of both sides :
log y = 31 log10 a
let
⇒
y =√
⎯3 a = a1/3
y = Antilog
[
1
3
log10 a
]
4·22
BUSINESS STATISTICS
COMPUTATION OF CUBE ROOTS
1
l og10 a
3
log 10 a
a
Anti-log [13 log10a ]
12·67
Ratio of sides
3·83 –~ 3·8
2031·9
3·3079
1·1026
354·0
2·5441
0·8480
7·047
81·6
1·9117
0·6372
4·337
62·2
1·7938
0·5979
3·962
2·13 ~
– 2·1
1·31 –~ 1·3
1·19 ~
– 1·2
36·3
1·5599
0·5199
3·311
1
Now we can express the data diagrammatically by drawing cubes with sides proportional to the values
given in the last column.
1
3
8 cms =
Scale :
2.1
Islam
3.8
Hinduism
36·3 lakhs
1·3
Christianity
1·2
Sikhism
1
Others
Fig. 4·23.
4·3·6. Pictograms. Pictograms is the technique of presenting statistical data through appropriate
pictures and is one of the very popular devices particularly when the statistical facts are to be presented to a
layman without any mathematical background. In this, the magnitudes of the particular phenomenon under
study are presented through appropriate pictures, the number of pictures drawn or the size of the pictures
being proportional to the values of the different magnitudes to be presented. Pictures are more attractive
and appealing to the eye and have a lasting impression on the mind. Accordingly they are extensively used
by government and private institutions for diagrammatic presentation of the data relating to a variety of
social, business or economic phenomena primarily for display to the general public or common masses in
fairs and exhibitions.
Remark. Pictograms have their limitations also. They are difficult and time-consuming to construct. In
pictogram, each pictorial symbol represents a fixed number of units like thousands, millions or crores, etc.
For instance, in Example 4·20 which displays the number of vessels in the Indian Merchant Shipping fleet,
one ship symbol represents 50 vessels and it is really a problem to represent and read fractions of 50. For
example, 174 vessels will be represented in pictograms by 3 ships and about a half more ; 231 vessels will
be represented in pictogram by 4 ships and a proportionate fraction of the 5th. This proportionate
representation introduces error and is quite difficult to visualise with precision. We give below some
illustrations of pictograms.
The following table gives the number of students studying in schools/colleges for different years in
India.
STUDENTS ON ROLL AT THE SCHOOL / UNIVERSITY STAGE
(As on 31st March)
Stage ↓
Class I to XI
University/College
1960-61
44·73
0·73
(In Million)
1965-66
1970-71
1974-75
1975-76
66·29
1·24
76·95
2·81
87·30
2·94
89·46
3·21
Source : Ministry of Education and Social Welfare.
4·23
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
The data can be represented in the diagrammatic form by pictogram (pictures) as given below :
STUDENTS ON ROLL AT SCHOOL/UNIVERSITY STAGE
CLASS I-XI
1960-61
= = 10 Million
1965-66
1970-71
1974-75
1975-76
UNIVERSITY/COLLEGE
= 0·5 Million
1960-61
1965-66
1970-71
1974-75
1975-76
Fig. 4·24.
Data pertaining to the population of a country at different age groups are usually represented through
pictures by means of so-called pyramids. The pyramid in Fig. 4·25 exhibits the population of India at
various age-groups according to 1971 census as given in the following table.
DISTRIBUTION OF POPULATION
BY AGE GROUPS (1971 Census)
Age-group
Under 1
1— 4
5—14
15—24
25—34
35—44
45—54
55—64
65—74
75 and over
Population (in ’000)
16,519
63,040
1,50,776
90,569
77,010
61,186
43,416
27,202
12,880
5,446
Source : Registrar General of India.
AGE PYRAMIDS — INDIA CENSUS 1971
75 and
Over
Population in Millions
65-74
55-64
45-54
35-44
25-34
15-24
5-14
1-4
0-1
7
6
5
4
3
2
1
0 1 2
Fig. 4·25.
3
4
5
6
7
8
4·24
BUSINESS STATISTICS
Example 4·20. The following table gives the number of vessels as on 31st December, in Indian
Merchant Shipping fleet for different years.
Year
No. of vessels
1961
174
1966
231
1971
255
1975
330
1976
359
Represent the data by pictogram.
Solution.
NUMBER OF VESSELS IN INDIAN MERCHANT SHIPPING FLEET
= 50 VESSELS
1961
1966
1971
1975
1976
Fig. 4·26.
4·3·7. Cartograms. In cartograms, statistical facts are presented through maps accompanied by various
types of diagrammatic representation. They are specially used to depict the quantitative facts on a regional
or geographical basis e.g., the population density of different states in a country or different countries in the
world, or the distribution of the rainfall in different regions of a country can be shown with the help of
maps or cartograms. The different regions or geographical zones are depicted on a map and the quantities
or magnitudes in the regions may be shown by dots, different shades or colours etc., or by placing bars or
pictograms in each region or by writing the magnitudes to be represented in the respective regions.
Cartograms are simple and elementary forms of visual presentation and are easy to understand. They are
generally used when the regional or geographic comparisons are to be highlighted.
4·3·8. Choice of a Diagram. In the previous sections we have described types of diagrams which can
be used to present the given set of numerical data and also discussed briefly their relative merits and
demerits. No single diagram is suited for all practical situations. The choice of a particular diagram for
visual presentation of a given set of data is not an easy one and requires great skill, intelligence and
expertise. The choice will primarily depend upon the nature of the data and the object of presentation, i.e.,
the type of the audience to whom the diagrams are to be presented and it should be made with utmost care
and caution. A wrong or injudicious selection of the diagram will distort the true characteristics of the
phenomenon to be presented and might lead to very wrong and misleading interpretations. Some special
types of data, viz., the data relating to frequency curves and time series are best represented by means of
graphs which we will discuss in the following sections.
EXERCISE 4·1
1. (a) What are the merits and limitations of diagrammatic representation of statistical data ?
(b) Describe the advantages of diagrammatic representation of statistical data. Name the different types of
diagrams commonly used and mention the situations where the use of each type of diagram would be appropriate.
2. (a) What are the different types of diagrams which are used in statistics to show the salient characteristics of
group and series ? Illustrate your answer.
(b) Discuss the usefulness of diagrammatic representation of facts.
4·25
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
3. “Diagrams do not add anything to the meaning of Statistics but when drawn and studied intelligently, they bring
to view the salient characteristics of the data”. Explain.
4. “Diagrams help us visualize the whole meaning of a numerical complex data at a single glance.” Explain the
statement.
[Delhi Univ. B.Com. (Pass), 1999]
5. The merits of diagrammatic presentation of data are classified under three heads : attraction, effective
impression and comparison. Explain and illustrate these points.
6. (a) State the different methods used for diagrammatic representation of statistical data and indicate briefly the
advantages and disadvantages of each one of them.
(b) Point out the usefulness of diagrammatic representation of facts and explain the construction of any of the
different forms of diagrams you know.
(c) A Bar diagram and a Rectangular diagram have the same appearance. Do they belong to the same category of
diagrams ? Explain.
[Delhi Univ. B.Com. (Pass), 1998]
7. What types of mistakes are commonly committed in the construction of diagrams ? What precautions are
necessary in this connection ?
8. (a) Draw a bar chart to represent the following information :
Year
No. of women M.P.’s
1952
22
1957
27
1962
34
1967
31
1972
22
1977
19
(b) In a recent study on causes of strikes in mills, an experimenter collected the following data.
Causes
:
Occurrences
(in percentage) :
Economic
Personal
Political
Rivalry
Others
58
16
10
6
10
Represent the data by bar chart.
9. Represent the following data by a percentage sub-divided bar-diagram :
Item of Expenditure
Family A (Income Rs. 500)
Family B (Income Rs. 300)
Food
150
Clothes
125
60
25
50
Education
150
Miscellaneous
190
70
Saving or Deficit
+10
–30
10. (a) Construct a multiple bar graph to represent Imports and Exports of a country for the following years :
Year
Imports (In billion rupees)
Exports (In billion rupees)
1992-93
19
20
1993-94
30
25
1994-95
45
33
1995-96
53
40
1996-97
51
51
(b) Draw a suitable diagram to present the following data :
1998
1999
I Division
II Division
III Division
Failures
Total No. of candidates
16
12
40
44
60
72
44
34
160
162
11. Represent the following data by a sub-divided bar diagram :
No. of Students
College
Arts
Science
Commerce
A
1200
800
600
B
750
500
300
Agriculture
400
450
Total
3000
2000
4·26
BUSINESS STATISTICS
12. Draw a rectangular diagram to represent the following information :
Factory A
Rs. 15·00
1000 Nos.
Rs. 5·00
Rs. 4·00
Rs. 6·00
Price per unit
Units produced
Raw material/unit
Other expenses/unit
Profit/unit
Factory B
Rs. 12·00
1200 Nos.
Rs. 5·00
Rs. 3·00
Rs. 4·00
13. Represent the following data by a deviation bar diagram :
Years
Income (in crores of Rs.)
Expenditure (in crores of Rs.)
1994
15
18
1995
16
17
1996
17
16
1997
18
20
1998
19
17
1999
20
18
[Delhi Univ. B.Com. (Pass), 2002]
14. (a) What do you mean by two-dimensional diagrams ? Under what situations they are preferred to onedimensional diagrams ?
(b) Describe the (i) square and (ii) circle diagrams. Discuss their merits and demerits.
(c) What do you understand by Pie diagrams. Discuss the technique of constructing such diagrams.
15. The following table gives the average approximate yield of rice in kg. per acre in three different countries.
Draw square diagrams to represent the data :
Country
Yield in (kg.) per acre
A
350
B
647
C
1120
16. Represent the following data on production of Tea, Cocoa and Coffee by means of a pie diagram.
Tea
3,260 tons
Cocoa
1,850 tons
Coffee
900 tons
Total
6,010 tons
17. (a) Point out the usefulness of diagrammatic representation of facts and explain the construction of volume and
pie diagrams.
(c) Draw a pie diagram for the following data of
(b) A Rupee spent on ‘Khadi’ is distributed as
Sixth Five-Year Plan Public Sector outlays:
follows :
Paise
Agriculture and Rural Development
12·9%
Farmer
19
Irrigation, etc.
12·5%
Carder and Spinner
35
Energy
27·2%
Weaver
28
Industry and Minerals
15·4%
Washerman, Dyer and Printer
8
Transport, Communication, etc.
15·9%
Administrative Agency
10
Social Services and Others
16·1%
Total 100
(Bangalore Univ. B.Com., 1997)
Present the data in the form of a pie diagram.
—————
—————
18. Draw a Pie diagram to represent the distribution of a certain blood group ‘O’ among Gypsies, Indians and
Hungarians.
Frequency
————————————————————————————————————————————————————————————————
Blood group
‘O’
Gypsies
343
Indians
313
19. (a) Represent the following data by means of circular diagrams.
No. of Employed
Year
Men
Women
1951
1,80,000
1,10,000
1961
3,50,000
2,10,000
Hungarians
344
Total
1000
Children
70,000
1,60,000
Total
3,60,000
7,20,000
4·27
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
(b) Represent the following data by Pie diagram.
Expenditure (in Rs.)
Items of Expenditure
Family A
Family B
Food
150
120
Clothing
100
80
Rent and Education
120
80
Fuel and Electricity
80
40
Others
90
40
20. The areas of the various continents of the world in millions of square miles are presented below :
AREAS OF CONTINENTS OF THE WORLD
Continent :
Area (Millions
of square miles) :
Africa
Asia
11·7
10·4
Europe North America Oceania South America
1·9
9·4
3·3
6·9
U.S.S.R
Total
7·9
51·5
Represent the data by a Pie diagram.
4·4. GRAPHIC REPRESENTATION OF DATA
–
–
–
–
–
–
–
–
–
The difference between the diagrams and graphs has been discussed in §4·2. To summarise, diagrams
are useful for visual presentation of categorical and geographical data while the data relating to time series
and frequency distributions is best represented through graphs. Diagrams are primarily used for
comparative studies and can’t be used to study the relationship, (not necessarily functional), between the
variables under study. This is done through graphs. Diagrams furnish only approximate information and are
not of much utility to a statistician from analysis point of view. On the other hand, graphs are more
obvious, precise and accurate than diagrams and can be effectively used for further statistical analysis, viz.,
to study slopes, rates of change and for forecasting wherever possible. Graphs are drawn on a special type
of paper, known as graph paper. The advantages of graphic representation of a set of numerical data have
also been discussed in § 4·1.
Like diagrams, a large number of graphs are used in practice. But they can be broadly classified under
the following two heads :
(i) Graphs of Frequency Distributions.
(ii) Graphs of Time Series.
Before discussing these graphs we shall briefly describe the technique of constructing graphs and the
general rules for drawing graphs.
Y
4·4·1. Technique of Construction of Graphs.
Graphs are drawn on a special type of paper known as
graph paper which has a fine net work of horizontal and
QUADRANT I
QUADRANT II +4 –
vertical lines; the thick lines for each division of a
+3 –
X-Positive
X-Negative
centimetre or an inch measure and thin lines for small
Y-Positive
Y-Positive
parts of the same. In a graph of any size, two simple
+2 –
(+x, +y)
(–x, +y)
lines are drawn at right angle to each other, intersecting
+1 –
at point ‘O’ which is known as origin or zero of
reference. The two lines are known as co-ordinate axes. X ′
–
X
1 2 3 4
– 4 –3 –2 –1 0
The horizontal line is called the X-axis and is denoted
–1 –
by X′OX. The vertical line is called the Y-axis and is
QUADRANT III
–2 – QUADRANT IV
usually denoted by YOY′. Thus, the graph is divided
X-Negative
X-Positive
into four sections, known as the four quadrants, but in
–3 –
Y-Negative
practice only the first quadrant is generally used unless
Y-Negative
–4 –
negative magnitudes are to be displayed. Along the X(–x, –y)
(+x, –y)
axis, the distances measured towards right of the origin
Y ′
i.e., towards right of the line YOY′, are positive and the
Fig. 4·27.
distances measured towards left of origin i.e., towards
left of the line YOY′ are negative, the origin showing
4·28
BUSINESS STATISTICS
–
–
–
–
–
–
–
–
–
the value zero. Along the Y-axis, the distances above
Y
the origin i.e., above the line X′OX are positive and
Q
(–3,
4)
the distances below the origin i.e., below the
+4 –
line X′OX are negative. Any pair of the values of the
variables is represented by a point (x, y), x usually
+3 –
P (4, 2)
represents the value of the independent variable and is
+2 –
shown along the X-axis and y represents the value of the
+1 –
dependent variable and is shown along the Y-axis. The
four quadrants along the position of x and y values are
–
X ′
X
shown in the Fig. 4·27.
– 4 –3 –2 –1 0
1 2 3 4
–1 –
In any pair (a, b), first coordinate, viz., ‘a’ always
refers to the X-coordinate which is also known as
–2 –
abscissa and the second coordinate, viz., ‘b’ always
R (3, –2)
–3 –
refers to the Y -coordinate which is also known as
S (–2, –3)
–4 –
ordinate. As an illustration the points P(4, 2), Q(–3, 4),
R(3, –2) and S(–2, –3) are displayed in the Fig. 4·28.
Y ′
In the graph on a natural or arithmetic scale, the
equal magnitudes of the values of the variables are
Fig. 4·28.
represented by equal distances along both the axes,
though the scales along X-axis and Y-axis may be different depending on the nature of the phenomenon
under consideration.
4·4·2. General Rules for Graphing. The following guidelines (some of which have already been
discussed in § 4·3·1 for diagrammatic representation of data), may be kept in mind for drawing effective
and accurate graphs :
1. Neatness. (For details see § 4·3·1 page 4.2).
2. Title and Footnote (For details see § 4·3·1 page 4.2).
3. Structural Framework. The position of the axes should be so chosen that the graph gives an
attractive and proportionate get up. It should be kept in mind that for each and every value of the
independent variable, there is a corresponding value of the dependent variable. In drawing the graph it is
customary to plot the independent variable along the X-axis and the dependent variable along the Y-axis.
For instance, if the data pertaining to the prices of the commodity and the quantity demanded or supplied at
different prices is to be plotted, then the dependent variable, viz., price (which depends on independent
forces of supply and demand) is taken along Y-axis while the independent variable viz., quantity demanded
or supplied is taken along X-axis. Similarly, in case of time series data, the time factor is taken along Xaxis and the phenomenon which changes with time e.g., population of a country in different years,
production of a particular commodity for different periods, etc., is taken along Y-axis.
4. Scale. This point has also been discussed in § 4·3·1. It may further be added that the scale along both
the axes (X-axis and Y-axis) should be so chosen that the entire data can be accommodated in the available
space without crowding. In this connection, it is worthwhile to quote the words of A. L. Bowley :
“It is difficult to lay down rules for the proper choice of scales by which the figures should be plotted
out. It is only the ratio between the horizontal and vertical scales that need to be considered. The figure
must be sufficiently small for the whole of it to be visible at once : if the figure is complicated, related to
long series of years and varying numbers, minute accuracy must be sacrificed to this consideration.
Supposing the horizontal scale is decided, the vertical scale must be chosen so that the part of the line
which shows the greatest rate of increase is well inclined to the vertical which can be managed by making
the scale sufficiently small; and on the other hand, all important fluctuations must be clearly visible for
which the scale may need to be decreased. Any scale which satisfies both these conditions will fulfill its
purpose.”
5. False Base Line. The fundamental principle of drawing graph is that the vertical scale must start
with zero. If the fluctuations in the values of the dependent variable (to be shown along Y-axis) are very
small relative to their magnitudes, and if the minimum of these values is very distant (far greater) from
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
4·29
zero, the point of origin, then for an effective portrayal of these fluctuations the vertical scale is stretched
by using false base line. In such a situation the vertical scale is broken and the space between the origin ‘O’
and the minimum value (or some convenient value near that) of the dependent variable is omitted by
drawing two zig-zag horizontal lines above the base line. The scale along Y-axis is then framed
accordingly. False base line technique is quite extensively used for magnifying the minor fluctuations in a
time series data. It also economises space because if such data are graphed without using false base line,
then the plotted data will lie on the top of the graph. This will give a very clumsy look and also result in
wastage of space. However, proper care should be taken to interpret graphs in which false base line is used.
As illustrations, see Examples 4·31 and 4·32 in § 4·4·4.
6. Ratio or Logarithmic Scale. In order to display proportional or relative changes in the magnitudes,
the ratio or logarithmic scale should be used instead of natural or arithmetic scale which is used to display
absolute changes. Ratio scale is discussed in detail in § 4·4·5.
7. Line Designs. If more than one variable is to be depicted on the same graph, the different graphs so
obtained should be distinguished from each other by the use of different lines, viz., dotted lines, broken
lines, dash-dot lines, thin or thick lines, etc., and an index to identify them should be given. [See Examples
4·31 and 4·32].
8. Sources Note and Number. For details see § 4·3·1.
9. Index. For details see § 4·3·1.
10. Simplicity. For details, see § 4·3·1.
Remark. For detailed discussion on items 1, 2, 4, 8, 9 and 10, see § 4·3·1 replacing the word ‘diagram’
by ‘graphs’.
4·4·3. Graphs of Frequency Distributions. The reasons and the guiding principles for the graphic
representation of the frequency distributions are precisely the same as for the diagrammatic and graphic
representation of other types of data. The so-called frequency graphs are designed to reveal clearly the
characteristic features of a frequency data. Such graphs are more appealing to the eye than the tabulated
data and are readily perceptible to the mind. They facilitate comparative study of two or more frequency
distributions regarding their shape and pattern. The most commonly used graphs for charting a frequency
distribution for the general understanding of the details of the data are :
(A) Histogram.
(B) Frequency Polygon.
(C) Frequency Curve.
(D) “Ogive” or Cumulative Frequency Curve.
The choice of a particular graph for a given frequency distribution largely depends on the nature of the
frequency distribution, viz., discrete or continuous. In the following sections we shall discuss them in
details, one by one.
A. HISTOGRAM
It is one of the most popular and commonly used devices for charting continuous frequency
distribution. It consists in erecting a series of adjacent vertical rectangles on the sections of the horizontal
axis (X-axis), with bases (sections) equal to the width of the corresponding class intervals and heights are
so taken that the areas of the rectangles are equal to the frequencies of the corresponding classes.
Construction of Histogram. The variate values are taken along the X-axis and the frequencies along
the Y-axis.
Case (i) Histogram with equal classes. In the case, if classes are of equal magnitude throughout, each
class interval is drawn on the X-axis by a section or base (of the rectangle) which is equal (or proportional)
to the magnitude of the class interval. On each class interval (as base) erect a rectangle with the height
proportional to the corresponding frequency of the class. The series of adjacent rectangles (one for each
class), so formed gives the histogram of the frequency distribution and its area represents the total
frequency of the distribution as distributed throughout the different classes. The procedure is explained in
Example 4.21.
Case (ii) Histogram with unequal classes. If all the classes are not uniform throughout; as in case (i)
the different classes are represented on the X-axis by sections or bases which are equal (or proportional) to
4·30
BUSINESS STATISTICS
the magnitudes of the corresponding classes and the heights of the corresponding rectangles are to be
adjusted so that the area of the rectangle is equal to the frequency of the corresponding class. This
adjustment can be done by taking the height of each rectangle proportional (equal) to the corresponding
frequency density of each class which is obtained on dividing the frequency of the class by its magnitude,
viz.,
Frequency of the class
Frequency Density (of a class) = Magnitude of the class .
Instead of finding the frequency density a more convenient way (from the practical point of view) is to
make all the class intervals equal and then adjust the corresponding frequency by using the basic
assumption that all the frequencies are distributed uniformly throughout the class. This consists in taking
the lowest class interval as standard one with unit length on the X-axis. The adjusted frequencies of the
different classes are obtained on dividing the frequency of the given class by the corresponding Adjustment
Factor (A.F.) which is given by :
Magnitude of the class
A.F. for any class = Lowest class interval .
Thus, if the magnitude of any class interval is twice (three) the lowest class interval, the adjustment
factor is 2(3) and the height of the rectangle which is represented by the adjusted frequency will be 12
( rd )
1
3
of the corresponding class frequency and so on. This is illustrated in Example 4·22. This adjustment gives
the rectangles whose areas are equal to the frequencies of the corresponding classes.
Remarks 1. Grouped (Not Continuous) Frequency Distribution. It should be clearly understood that
histogram can be drawn only if the frequency distribution is continuous. In case of grouped frequency
distribution, if classes are not continuous, they should be made continuous by changing the class limits into
class boundaries and then rectangles should be erected on the continuous classes so obtained. As an
illustration, see Example 4·24.
2. Mid-points given. Sometimes, only the mid-values of different classes are given. In such a case, the
given distribution is converted into continuous frequency distribution with exclusive type classes by
ascertaining the upper and lower limits of the various classes under the assumption that the class
frequencies are uniformly distributed throughout each class. (See Example 4·25.)
3. Discrete Frequency Distribution. Histograms, may sometimes also be used to represent discrete
frequency distribution by regarding the given values of the variable as the mid-points of continuous classes
and then proceeding as explained in Remark 2 above.
4. Difference between Histogram and Bar Diagram. (i) A histogram is a two-dimensional (area)
diagram where both the width (base) and the length (height of the rectangle) are important whereas bar
diagram is one-dimensional diagram in which only length (height of the bar) matters while width is
arbitrary.
(ii) In a histogram, the bars (rectangles) are adjacent to each other whereas in bar diagram proper
spacing is given between different bars.
(iii) In a histogram, the class frequencies are represented by the area of the rectangles while in a bar
diagram they are represented by the heights of the corresponding bars.
5. Open-end classes. Histograms can’t be constructed for frequency distributions with open end classes
unless we assume that the magnitude of the first open class is same as that of the succeeding (second) class
and the magnitude of the last open class is same as that of the preceding (i.e., last but one) class.
6. Histogram may be used for the graphic location of the value of Mode (See Chapter 5).
Example 4·21. Draw histogram for the following frequency distribution.
Variable
Frequency
:
:
10—20
12
20—30
30
30—40
35
40—50
65
50—60
45
60—70
25
70—80
18
4·31
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
Solution.
HISTOGRAM
70
65
60
FREQUENCY
50
45
40
35
30
30
25
18
20
12
10
0
10
20
30
40
50
60
70
80
90
VARIABLE
Fig. 4·29.
Example 4·22. Represent the following data by means of a histogram.
Weekly Wages (’00 Rs.) :
No. of Workers
:
10—15
7
15—20
19
20—25
27
25—30
15
30—40
12
40—60
12
60—80
8
No. of
Magnitude
Height of
Solution. Since the class intervals are of Weekly Wages
Workers (f)
of Class
Rectangle
(’00 Rs.)
unequal magnitude, the corresponding
frequencies have to be adjusted to obtain the
10—15
7
5
7
so-called ‘frequency density’ so that the area
15—20
19
5
19
of the rectangle erected on the class interval
20—25
27
5
27
is equal to the class frequency. We observe
25—30
15
5
15
that first four classes are of magnitude 5, the
30—40
12
10
(12/2) = 6
class 30—40 is of magnitude 10 and the last
40—60
12
20
(12/4) = 3
two classes 40—60 and 60—80 are of
60—80
8
20
(8/4) = 2
magnitude 20. Since 5 is the minimum class
interval, the frequency of the class 30—40 is divided by 2 and the frequencies of classes 40—60 and 60—
80 are to be divided by 4 as shown in the adjoining table.
——————————————————————————————————————————— —————————————————————
—————————————————————
HISTOGRAM
27
NUMBER OF WORKERS
30
25
19
20
15
15
10
7
5
0
6
3
2
60
10 15 20 25 30
70
40 50
WEEKLY WAGES (IN ’00 Rs.)
Fig. 4·30.
80
4·32
BUSINESS STATISTICS
B. FREQUENCY POLYGON
Frequency polygon is another device of graphic presentation of a frequency distribution (continuous,
grouped or discrete).
In case of discrete frequency distribution, frequency polygon is obtained on plotting the frequencies on
the vertical axis (Y-axis) against the corresponding values of the variable on the horizontal axis (X-axis)
and joining the points so obtained by straight lines. [As an illustration, see Example 4·23.]
In case of grouped or continuous frequency distribution, frequency polygon may be drawn in two
ways.
Case (i) From Histogram. First draw the histogram of the given frequency distribution as explained in
§ 4·4·3 (A). Now join the mid-points of the tops (upper horizontal sides) of the adjacent rectangles of the
histogram by straight line graph. The figures so obtained is called a frequency polygon. (Polygon is a figure
with more than four sides). It may be noted that when the frequency polygon is constructed as explained
above it cuts off a triangular strip (which lies outside the frequency polygon) from each rectangle of the
histogram. But, at the same time, another triangular strip of the same area which is outside the histogram is
included under the polygon, as shown by shaded area in the diagram of Example 4·28. [This, however, is
not true in the case of unequal class intervals]. In order that the area of the frequency polygon is equal to
the area of the corresponding histogram of the frequency distribution, it is necessary to close the polygon at
both ends by extending them to the base line such that it meets the X-axis at the mid-points of two
hypothetical classes, viz., the class before the first class and the class after the last class, at both the ends
each with frequency zero [See Examples 4·24 and 4·25].
Case (ii) Without Constructing Histogram. Frequency polygon of a grouped or continuous frequency
distribution is a straight line graph which can also be constructed directly without drawing the histogram.
This consists in plotting the frequencies of different classes (along Y-axis) against the mid-values of the
corresponding classes (along X-axis). The points so obtained are joined by straight lines to obtain the
frequency polygon. As in Case (i), the frequency polygon so obtained should be extended to the base at
both ends by joining the extreme points (first and last point) to the mid-points of the two hypothetical
classes (before the first class and after the last class) assumed to have zero frequencies. The figure of the
frequency polygon so obtained would be exactly same as in Case (i) except for the histogram.
This point can be elaborated mathematically as follows. Let x1, x2, …, x n be the mid-values of n classes
with frequencies f1 , f2, …, fn respectively. We plot the points (x1, f1), (x 2 , f2),…, (xn, fn) on the co-ordinate
axes, taking mid-values along X-axis and frequencies along Y-axis and join them by straight lines. The first
point (x1 , f1 ) is joined to the point (x0, 0) and the last point (xn, fn ) to the point (x n + 1, 0) by straight lines and
the required frequency polygon is obtained.
Remarks 1. Frequency polygon can be drawn directly without the histogram (as explained above) if
only the mid-points of the classes are given; without forming the continuous frequency distribution which is
desirable in the case of histogram.
2. Frequency Polygon Vs. Histogram : (i) Histogram is a two-dimensional figure, viz., a collection of
adjacent rectangles whereas frequency polygon is a line graph.
(ii) Frequency polygon can be used more effectively for comparative study of two or more frequency
distributions because frequency polygons of different distributions can be drawn on the same single graph.
This is not possible in the case of histogram where we need separate histograms for each of the frequency
distributions. However, for studying the relationship of the individual class frequencies to the total
frequency, histogram gives a better picture and is accordingly preferred to the frequency polygon.
(iii) In the construction of frequency polygon we come across same difficulties as in the construction of
histograms, viz.,
(a) It cannot be constructed for frequency distributions with open end classes; and
(b) Suitable adjustments, as in the case of histogram are required for frequency distributions with
unequal classes.
4·33
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
(iv) Unlike histogram, frequency polygon is a continuous curve and therefore possesses all the distinct
advantages of graphic representation, viz., it may be used to determine the slope, rate of change, estimates
(interpolation and extrapolation), etc., wherever admissible.
Example 4·23. The following data show the number of accidents sustained by 313 drivers of a public
utility company over a period of 5 years.
Number of accidents :
0
1
2
3
4
5
6
7
8
9
10
11
Number of drivers
80
44
68
41
25
20
13
7
5
4
3
2
Draw the frequency polygon.
Solution. See Fig. 4·31
FREQUENCY POLYGON OF NUMBER OF ACCIDENTS
NUMBER OF DRIVERS
80
70
FREQUENCY
POLYGON
60
50
40
30
20
10
0
1
2
3
4
5
6 7
8
NUMBER OF ACCIDENTS
9
10 11
Fig. 4·31.
Example 4·24. The following table gives the frequency distribution of the weekly wages (in ’00 Rs.) of
100 workers in a factory.
Weekly Wages
(’00 Rs.)
Number of Workers
20—24
25—29
30—34
35—39
40—44
45—49
50—54
55—59
60—64
Total
4
5
12
23
31
10
8
5
2
100
Draw the histogram and frequency polygon of the distribution.
Solution. Since all the classes are of equal magnitude i.e., 5, for the construction of the histogram, the
heights of the rectangles to be erected on the classes will be proportional to their respective frequencies.
However, since the classes are not continuous, the given distribution is to be converted into a continuous
frequency distribution, with exclusive type classes before erecting the rectangles, as given in the following
table.
Weekly Wages
19·5—24·5 24·5—29·5 29·5—34·5 34·5—39·5 39·5—44·5 44·5—49·5 49·5—54·5 54·5—59·5 59·5—64·5
(’00 Rs.)
Number of
4
5
12
23
31
10
8
5
2
Workers (f)
As usual, frequency polygon is obtained from histogram by joining the mid-points of the rectangles by
straight lines, and extended both ways to the classes 14·5—19·5 and 64·5—69·5 on the X-axis, as shown in
the Fig. 4·32.
4·34
BUSINESS STATISTICS
HISTOGRAM AND FREQUENCY POLYGON
31
NUMBER OF WORKERS
35
30
HISTOGRAM
25
23
20
POLYGON
15
12
10
10
8
4
5
5
5
2
0
14·5 19·5 24·5 29·5 34·5 39·5 44·5 49·5 54·5 59·5 64·5 69·5
WEEKLY WAGES
Fig. 4·32.
Remark. It may be pointed out that frequency polygon can be drawn straight way by plotting the
frequencies against the mid-points of the corresponding classes without converting the given distribution
into a continuous one and joining these points by straight lines.
Example 4·25. Draw the histogram and frequency polygon for the following frequency distribution.
:
:
2·5
7
7·5
10
Solution. Since we are given the
mid-values of the class intervals, for the
construction of histogram, the distribution
is to be transformed into continuous class
intervals each of magnitude 5, (under the
assumption that the frequencies are
uniformly distributed throughout the class
intervals), as given in the following table.
Class
0—5
5—10
10—15
15—20
20—25
25—30
30—35
35—40
Frequency
7
10
20
13
17
10
14
9
12·5
20
17·5
13
22·5
17
27·5
10
32·5
14
37·5
9
HISTOGRAM AND FREQUENCY POLYGON
25
20
FREQUENCY
Mid-value of class interval
Frequency
Histogram
15
Frequency
Polygon
10
5
0
5
10
15
20
25
30
35
40
45
The histogram and frequency polygon
CLASSES
are shown in the Fig. 4·33.
Fig. 4·33.
Remark. It may be noted that frequency polygon can be drawn even without converting the given
distribution into classes. The frequencies are plotted against the corresponding mid-points (given) and
joined by straight lines.
C. FREQUENCY CURVE
A frequency curve is a smooth free hand curve drawn through the vertices of a frequency polygon. The
object of smoothing of the frequency polygon is to eliminate, as far as possible, the random or erratic
fluctuations that might be present in the data. The area enclosed by the frequency curve is same as that of
4·35
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
the histogram or frequency polygon but its shape is smooth one and not with sharp edges. Frequency curve
may be regarded as a limiting form of the frequency polygon as the number of observations (total
frequency) becomes very large and the class intervals are made smaller and smaller.
Remarks 1. Smoothing should be done very carefully so that the curve looks as regular as possible
and sudden and sharp turns should be avoided. In case of the data pertaining to natural phenomenon like
tossing of a coin or throwing of a dice the smoothing can be conveniently done because such data generally
give rise to symmetrical curves. However, for the data relating to social, economic or business
phenomenon, smoothing cannot be done effectively as such data usually give rise to skewed (asymmetrical)
curves. [For details see § 4·4·3 C (b) page 4·36]. In fact, it is desirable to attempt a frequency curve if
we have sufficient reasons to believe that the frequency distribution under study is fairly regular. It is
futile to attempt a frequency curve for an irregular distribution. In general, frequency curves should be
attempted
(i) for frequency distribution based on the samples, and
(ii) when the distribution is continuous.
2. We have already seen that a frequency polygon can be drawn with or without a histogram.
However, to obtain an ideal frequency curve for a given frequency distribution, it is desirable to proceed in
a logical sequence, viz., first draw a histogram, then a frequency polygon and finally a frequency curve,
because in the absence of a histogram the smoothing of the frequency polygon cannot be done properly.
As discussed in frequency polygon, the frequency curve should also be extended to the base on both sides
of the histogram so that the area under the frequency curve represents the total frequency of the
distribution.
3. A frequency curve can be used with advantage for interpolation [i.e., estimating the frequencies for
given value of the variable or in a given interval (within the given range of the variable)], provided it rises
gradually to the highest point and then falls more or less in the same manner. It can also be used to
determine the rates of increase or decrease in the frequencies. It also enables us to have an idea about the
Skewness and Kurtosis of the distribution [See Chapter 7].
Example 4·26. Draw a frequency curve for the following distribution :
Age (Yrs.)
No. of Students
:
:
17—19
7
19—21
13
21—23
24
23—25
30
25—27
22
27—29
15
29—31
6
Solution. See Fig. 4·34.
NUMBER OF STUDENTS
FREQUENCY CURVE
30
Frequency Curve
25
Histogram
20
15
10
5
15
17
19
23 25 27 29
AGE (YEARS)
Fig. 4·34.
21
31
33
Types of Frequency Curves. Though different types of data may give rise to a variety of frequency
curves, we shall discuss below only some of the important curves which, in general, describe most of the
data observed in practice, viz., the data relating to natural, social, economic and business phenomena.
(a) Curves of Symmetrical Distributions. In a symmetrical distribution, the class frequencies first
rise steadily, reach a maximum and then diminish in the same identical manner.
4·36
BUSINESS STATISTICS
If a curve is folded symmetrically about a vertical line (corresponding to the maximum frequency), so
that the two halves of the figures coincide, it is called a symmetrical curve. It has a single smooth hump in
the middle and tapers off gradually at either end and is bell-shaped.
The following hypothetical distribution of marks in a test will give a symmetrical frequency
distribution.
Marks
Frequency
:
:
0—10
40
10—20
70
20—30
120
30—40
160
40—50
180
50—60
160
60—70
120
70—80
70
80—90
40
If the data are presented graphically, we shall obtain a frequency curve which is symmetrical.
The most commonly and widely used symmetrical curve in Statistics is the Normal frequency curve
which is given in Fig. 4·35. (For details, see Chapter 14 on Theoretical Probability Distributions).
NORMAL PROBABILITY CURVE
Normal curve, generally describes the data relating to
natural phenomenon like tossing of a coin, throwing of a
dice, etc. Most of the data relating to psychological and
educational statistics also give rise to normal curve.
However, the data relating to social, business and economic
phenomena do not conform to normal curve. They always
X = Mean
give moderately asymmetrical (slightly skewed) curves
Fig.
4·35.
discussed below.
(b) Moderately Asymmetrical (Skewed) Frequency Curves. A frequency curve is said to be skewed
(asymmetrical) if it is not symmetrical. Moderately asymmetrical curves are commonly observed in social,
economic and business phenomena. Such curves are stretched more to one side than to the other. If the
curve is stretched more to the right (i.e., it has a longer tail towards the right), it is said to be positively
skewed and if it is stretched more to the left (i.e., has a longer tail towards the left), it is said to be
negatively skewed. Thus, in a positively skewed distribution, most of the frequencies are associated with
smaller values of the variable and in a negatively skewed distribution most of the frequencies are associated
with larger values of the variable. The following figures show positively skewed and negatively skewed
distributions.
POSITIVELY SKEWED
DISTRIBUTION
NEGATIVELY SKEWED
DISTRIBUTION
Mode
Mode
Fig. 4·36.
(c) Extremely Asymmetrical or J-Shaped Curves. The distributions in which the value of the
variable corresponding to the maximum frequency is at one of the ranges (and not in the middle as in the
case of symmetrical distributions), give rise to highly skewed curves. When plotted, they give a J-shaped or
inverted J-shaped curve and accordingly such curves are also called J-shaped curves. In a J-shaped curve,
the distribution starts with low frequencies in the lower classes and then frequencies increase steadily as the
variable value increases and finally the maximum frequency is attained in the last class thus exhibiting a
peak at the extreme right end of the distribution. Such curves are not regular curves but become
unavoidable in certain situations. For example, the distribution of mortality (death) rates (along Y-axis)
w.r.t. age (along X-axis) after ignoring the accidental deaths; or the distribution of persons travelling in
local state buses, (e.g., DTC in Delhi or BEST in Mumbai) w.r.t. time from morning hours, say, 7 A.M. to
peak traffic hours, say, 10 A.M. will give rise to a J-shaped distribution [Fig. 4·37(a)]. Similarly in an
inverted J-shaped curve the frequency decreases continuously with the increase in the variate values, the
maximum frequency being attained in the beginning of the distribution. For example, the distribution of the
quantity demanded w.r.t. the price; or the number of depositors w.r.t. their saving in a bank, or the number
of persons w.r.t. their wages or incomes in a city, will give a reverse J-shaped curve [Fig. 4·37(b)].
4·37
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
INVERTED J-SHAPED CURVE
Mortality Rate
Quantity Demanded
J-SHAPED CURVE
Price
Age
Fig. 4·37(a).
Fig. 4·37(b).
No. of persons
(d) U-Curve. The frequency distributions in which
U-SHAPED CURVE
the maximum frequency occurs at the extremes (i.e., both f
ends) of the range and the frequency keeps on falling
symmetrically (about the middle), the minimum
frequency being attained at the centre give rise to a Ushaped curve. In this type of distribution, most of values
are associated with the values of the variable at the
extremes i.e., with smaller and larger values whereas
smaller frequencies are associated with the intermediate
values, the central value having the minimum frequency.
Such distributions are generally observed in the
behaviour of total costs where the curve initially falls O Morning
Evening
Time
steadily and after attaining the optimum level (in the
peak hours
peak hours
middle), it starts rising steadily again. As another
Fig. 4·38.
illustration, the distribution of persons travelling in local
state buses between morning and evening peak hours will give, more or less, a U-shaped curve shown in
Fig. 4·38.
(e) Mixed Curves. In the curves discussed so far, we have seen that the highest concentration of the
values lies at the centre (symmetrical curve), or near around the centre (moderately asymmetrical curve), or
at the extremes (J-shaped and U-shaped curves). But sometimes, though very rarely, we come across
certain distributions in which maximum frequency is attained at two or more points in an irregular manner
as shown in Fig. 4·39.
Y
Bi-Modal Curve
O
Variable
Frequency
Frequency
Y
X
O
Tri-Modal Curve
Variable
X
Fig. 4·39.
Such curves are obtained in a distribution where as the value of the variable increases, the frequencies
increase and decrease, then again increase and decrease in an irregular manner; the phenomenon may be
repeated twice or thrice as shown in the above diagrams or even more than that. The distributions with two
humps are called bi-modal distributions and those with three humps are called tri-modal distributions while
those with more than three humps are termed as multi-modal distributions. Such distributions are rarely
observed in practice and should be avoided as far as possible because they cannot be usefully employed for
the computation of various statistical measures and for statistical analysis.
4·38
BUSINESS STATISTICS
D. OGIVE OR CUMULATIVE FREQUENCY CURVE
Ogive, pronounced as ojive, is a graphic presentation of the cumulative frequency (c.f.) distribution
[See Chapter 3] of continuous variable. It consists in plotting the c.f. (along the Y-axis) against the class
boundaries (along X-axis). Since there are two types of cumulative frequency distributions viz., ‘less than’
c.f. and ‘more than’ c.f. we have accordingly two types of ogives, viz.,
(i) ‘Less than’ ogive.
(ii) ‘More than’ ogive.
‘Less Than’ Ogive. This consists in plotting the ‘less than’ cumulative frequencies against the upper
class boundaries of the respective classes. The points so obtained are joined by a smooth freehand curve to
give ‘less than’ ogive. Obviously, ‘less than’ ogive is an increasing curve, sloping upwards from left to
right and has the shape of an elongated S.
Remark. Since the frequency below the lower limit of the first class (i.e., upper limit of the class
preceding the first class) is zero, the ogive curve should start on the left with a cumulative frequency zero
at the lower boundary of the first class.
‘More Than’ Ogive. Similarly, in ‘more than’ ogive, the ‘more than’ cumulative frequencies are
plotted against the lower class boundaries of the respective classes. The points so obtained are joined by a
smooth freehand curve to give ‘more than’ ogive. ‘More than’ ogive is a decreasing curve and slopes
downwards from left to right and has the shape of an elongated S, upside down.
Remarks 1. We may draw both the ‘less than’ ogive and ‘more than’ ogive on the same graph. If done
so, they intersect at a point. The foot of the perpendicular from their point of intersection on the X-axis
gives the value of median. [See Example 4·28].
2. Ogives are particularly useful for graphic computation of partition values, viz., Median, Quartiles,
Deciles, Percentiles, etc. [For details, see Chapter 5]. They can also be used to determine graphically the
number or proportion of observations below or above a given value of the variable or lying between certain
interval of the values of the variable.
3. Ogives can be used with advantage over frequency curves for comparative study of two or more
distributions because like frequency curves, for each of the distributions different ogives can be constructed
on the same graph and they are generally less overlapping than the corresponding frequency curves.
4. If the class frequencies are large, they can be expressed as percentages of the total frequency. The
graph of the cumulative percentage frequency is called ‘percentile curve’.
Example 4·27. Draw a less than cumulative frequency curve for the following data and find from the
graph the value of seventh decile.
Monthly income
No. of workers
0—100
100—200
200—300
300—400
400—500
12
28
35
65
30
Monthly income
500—600
600—700
700—800
800—900
900—1000
No. of workers
20
20
17
13
10
Solution. Less than cumulative frequency curve is obtained on plotting the ‘less than’ c.f. against the
upper limit of the corresponding class and joining the points so obtained by a smooth free hand curve as
shown in Fig. 4·40.
7N 7
To obtain the value of seventh decile from the graph, at frequency
=
× 250 = 175, draw a line
10 10
parallel to the X-axis meeting the ‘less than’ c.f. curve at point P. From P draw PM perpendicular to X-axis
meeting it at M. Then the value of seventh decile is (Fig. 4.40) :
D7 = OM = Rs. 545 [From the graph]
4·39
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
——————————————————————
——————————————————————
————————————————————
250
NO. OF WORKERS
‘LESS THAN’ CUMULATIVE
FREQUENCY TABLE
Monthly
No. of
Less than
Income
workers (f)
c.f.
0—100
12
12
100—200
28
40
200—300
35
75
300—400
65
140
400—500
30
170
500—600
20
190
600—700
20
210
700—800
17
227
800—900
13
240
900—1000
10
250
200
P
150
7N
10
100
50
O
M 545
0
———————————————————————
—————————————————————————————————————
Total
∑ f = 250
100 200 300 400 500 600 700 800 900 1000
MONTHLY INCOME
Fig. 4·40.
Example 4·28. The following table gives the distribution of monthly income of 600 families in a certain
city.
Monthly Income (’00 Rs.)
No. of Families
Below 75
60
75—150
170
150—225 225—300
200
60
300—375 375—450
50
40
450 and over
20
Draw a ‘less than’ and a ‘more than’ ogive curve for the above data on the same graph and from these
read the median income.
Solution. For drawing the ‘less than’ and ‘more than’ ogive we convert the given distribution into ‘less
than’ and ‘more than’ cumulative frequencies (c.f.) as given in the following table.
No. of Families (f)
60
170
200
60
50
40
20
As already explained, for drawing ‘less than’
ogive, we plot ‘less than’ c.f. against the upper limit
of the corresponding class intervals and join the points
so obtained by smooth freehand curve. Similarly,
‘more than’ ogive is obtained on joining the points
obtained on plotting the ‘more than’ c.f. against the
lower limit of the corresponding class by smooth
freehand curve.
From the point of intersection of these two
ogives, draw a line perpendicular to the X-axis
(monthly incomes). The abscissa (x-coordinate) of the
point where this perpendicular meets the X-axis gives
the value of median.
The ‘more than’ and ‘less than’ ogives and the
value of median are shown in the Fig. 4·41.
From Fig. 4·41, Median = OM = 176
(approximately)
Hence, the median monthly income is Rs. 17,600.
Less than c.f.
60
230
430
490
540
580
600
More than c.f.
600
540
370
170
110
60
20
‘LESS THAN’ AND ‘MORE THAN’ OGIVE
NUMBER OF FAMILIES
Monthly Income (’00Rs.)
Below 75
75—150
150—225
225—300
300—375
375—450
450 and over
600
Less than
Ogive
500
400
300
200
100
M
O
More than
Ogive
0 50 100 150 200 250 300
400
MEDIAN = 176 (APPROX.)
MONTHLY INCOME (RS.)
Fig. 4·41.
500
4·40
BUSINESS STATISTICS
Example 4·29. Draw a percentile curve for the following distribution of marks obtained by 700
students at an examination.
Marks
No. of Students
Marks
No. of Students
0—10
9
50—59
102
10—19
42
60—69
71
20—29
61
70—79
23
30—39
140
80—89
2
40—49
250
Find from the graph
(i) the marks at the 20th percentile, and (ii) the percentile equivalent to a mark of 65.
Solution. A percentile curve is obtained on expressing the ‘less than’ cumulative frequencies as
percentage of the total frequency and then plotting these cumulative percentage frequencies (P) against the
upper limit of the corresponding class boundaries (x). These points are then joined by a smooth freehand
curve.
COMPUTATION OF CUMULATIVE PERCENTAGE FREQUENCY DISTRIBUTION
‘Less than’ c.f.
9
42
61
140
250
102
71
23
2
1·3
7·3
16·0
36·0
71·7
86·3
96·4
99·7
100·0
PERCENTILE CURVE
Y
100
B
90
80
70
60
50
99·5
89·5
M
79·5
A
10
O0
59·5
20
69·5
40
30
39·5
49·5
Percentage ‘less than’ c.f.
‘Less than’ c.f.
=
× 100
Total frequency
c.f.
c.f.
=
× 100 =
700
7
(i) To find marks (x) corresponding to the
20th percentile, at P = 20, draw a line
parallel to X-axis, meeting the percentile
curve at A. Draw AM perpendicular to Xaxis, meeting X-axis at M. Then
OM
= 31·5, gives the marks at the 20th
percentile.
(ii) To find percentile equivalent to mark
x = 65, at x = 65, draw perpendicular to Xaxis meeting the percentile curve at B.
From B draw a line parallel to X-axis
meeting the Y-axis at N. Then ON = 92, is
the percentile equivalent to score of 65.
Percentage ‘less than’ c.f. (P)
9
51
112
252
502
604
675
698
700
29·5
Frequency (f)
9·5
19·5
— 9·5
9·5—19·5
19·5—29·5
29·5—39·5
39·5—49·5
49·5—59·5
59·5—69·5
69·5—79·5
79·5—89·5
PERCENTAGE LESS THAN c.f. → P
Marks
X
MARKS
Fig. 4·42.
4·4·4. Graphs of Time Series or Historigrams. A time series is an arrangement of statistical data
in a chronological order i.e., with respect to occurrence of time. The time period may be a year, quarter,
month, week, days, hours and so on. Most of the series relating to economic and business data are time
series such as population of a country, money in circulation, bank deposits and clearings, production and
price of commodities, sales and profits of a departmental store, imports and exports of a country, etc. Thus
in a time series data there are two variables; one of them, the independent variable being time and the other
(dependent) variable being the phenomenon under study.
4·41
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
The time series data are represented geometrically by means of Time Series Graph which is also
known as Historigram. The independent variable viz., time is taken along the X-axis and the dependent
variable is taken along the Y-axis. The various points so obtained are joined by straight lines to get the time
series graph. If the actual time series data are graphed, the historigram is called Absolute Historigram.
However, the graph obtained on plotting the index number of the given values is called Index Historigram
and it depicts the percentage changes in the values of the phenomenon as compared to some fixed base
period. Historigrams are extensively used in practice. They are easy to draw and understand and do not
require much skill and expertise to construct and interpret them.
Remark. Time series graphs can be drawn on a natural (arithmetic scale) or on a ratio (semilogarithmic or logarithmic) scale, the former reflecting the absolute changes from one period to another and
the latter depicting the relative changes or rates of change. In the following sections we shall study the time
series graphs on a natural scale. Ratio scale graphs are discussed in § 4·4·5.
The various types of time series graphs are :
(i) Horizontal Line Graphs or Historigrams
(ii) Silhouette or Net Balance Graphs
(iii) Range or Variation Graphs
(iv) Components or Band Graphs
Now we shall discuss them briefly, one by one.
A. HORIZONTAL LINE GRAPHS OR HISTORIGRAMS
In such a graph only one variable is to be represented graphically. As already explained, the desired
graph (historigram) is obtained on plotting the time variable along the X-axis and the other variable viz., the
magnitudes of the phenomenon under consideration along the Y-axis on a suitable scale and joining the
points so obtained by straight lines. An illustration is given in Example 4·37.
Example 4·30. Draw the graph of the following :
1990
12·8
1991
13·9
1992
12·8
Solution. Taking the scale along X-axis as 1
cm = 1 year and along Y-axis as 1 cm = 2
million tons, the required graph is as shown in
Fig. 4·43.
FALSE BASE LINE. As already explained
[§ 4·4·2 (Item 5)], if the fluctuations in the
values of the variable (to be shown along the Yaxis) are small as compared to their magnitudes
and if the minimum value of the variable is very
distant from origin i.e., zero, then the technique
of false base line is used to highlight these
fluctuations. Illustrations are given in Examples
4·31 and 4·32.
1993
13·9
1994
13·4
1995
6·5
1996
2·9
1997
14·8
YIELD (IN MILLION TONS) FOR DIFFERENT YEARS
16
YIELD (IN MILLION TONS)
Year
Yield (in million tons)
14
12
10
8
6
4
2
1990 1991 1992 1993 1994 1995 1996 1997
YEARS
Fig. 4·43.
HISTORIGRAM — TWO OR MORE VARIABLES
The time series data relating to two or more related variables i.e., phenomena measured in the same
unit and belonging to the same time period can be displayed together in the same graph using the same
scales for all the variables along the vertical axis and the same scale for time along X-axis for each variable.
The method for drawing such graphs is same as that of historigram for one variable. Thus we shall get a
4·42
BUSINESS STATISTICS
number of curves, one for each variable. They should be distinguished from each other by the use of
different types of lines viz., thin and thick lines, dotted lines, dash lines, dash-dot lines, etc., and an index to
this effect should be given for proper identification of the curves. The following illustration will clarify the
point.
Example 4·31. The following table gives the index numbers of industrial production for India.
INDEX NUMBER OF INDUSTRIAL PRODUCTION
Item
Cement
Iron and steel
General Index
1971
107·0
100·6
104·2
1972
113·1
112·0
110·2
1973
107·6
96·1
112·0
1974
102·6
100·2
114·3
1975
116·7
121·3
119·3
Base : 1970 = 100
1976
133·9
145·0
131·2
Represent them on the same graph paper.
Ans. As usual, we take time (years) along X-axis and the index numbers along Y-axis. Using false base
line (for vertical axis) at 95, the graph is shown in Fig. 4·44.
150
INDEX NUMBERS
140
INDEX NUMBER OF INDUSTRIAL PRODUCTION
Base : 1970 = 100
Cement
Iron & Steel
General Index
130
120
110
100
95
0 1971
1972
1973
1974
YEARS
1975
1976
Fig. 4·44.
Remarks 1. The technique of drawing two or more historigrams on the same graph facilitates
comparisons between the related phenomena. However, its use should not be recommended if the number
of variables is large, say, more than 4. In such a case the different line graphs which may intersect each
other become quite confusing and it becomes quite difficult to understand and interpret them.
2. The graph obtained on plotting the index number is known as index historigram and it represents the
relative changes in the values of the variables under consideration. Two or more variable index
historigrams on the same graph obviously facilitate comparisons. However, in order to arrive at any valid
conclusion, the index numbers for all the variables should be computed with respect to the same base
period.
3. Graphs of two variables measured in different units. The time series data relating to two related
phenomena which are measured in different units e.g., imports (quantity in million tons) and imports
(values in crores Rupees) but pertaining to the same time period can also be displayed on the same graph.
This is done by using two different vertical scales (one for each variable), one on the left and the other on
4·43
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
the right; the scales for each variable being so selected that the two historigrams so obtained are close
to each other. This objective can be achieved by taking the scales for each variable proportional to its
average value i.e., the average value of each variable is kept in or near about the middle of the vertical scale
in the graph and the scale for each is selected accordingly. We explain this point by the following
illustration.
Example 4·32. Plot a graph to represent the following data in a suitable manner.
1990
400
220
1991
450
235
1992
560
385
1993
1994
1995
1996
1997
620
580
460
500
540
420
420
380
360
400
VOLUME AND VALUE OF IMPORTS
(1990—1997)
Volume
Solution. The time variable (Year) is recorded
along the X-axis with scale 1 cm = 1 year and the
variate values, imports (volume in tons) and imports
(value in Rs.) are recorded along the Y-axis with
scales :
1 cm = 20,000 tons (for imports-quantity)
1 cm = 20,000 Rs. (for imports-value)
and false base line is selected at 400,000 tons (for
quantity) and Rs. 220,000 (for value). The graph is
shown in Fig. 4·45.
IMPORTS (MILLION TONS)
620
600
Value
420
580
400
560
380
540
360
520
500
340
Volume
300
480
460
320
Value
IMPORTS (MILLION RS.)
Year
Imports (million tons)
Imports (million Rs.)
280
440
260
420
240
400
220
0
1990 1991 1992 1993 1994 1995 1996 1997
YEARS
Fig. 4·45.
B. SILHOUETTE OR NET BALANCE GRAPH
This graph is specially used to highlight the difference or the net balance between the values of two
variables along the vertical axis e.g., the difference between imports and exports of a country in different
years, sales and purchase of a business concern for different periods, the income and expenditure of a
family in different months and so on. This can be done in any one of the following two ways :
Method 1. Obtain the net balance viz., the difference between two sets of the values of the phenomena
(variables) for different periods. Some of these differences may be negative also. Now, in addition to the
two historigrams, one for each variable, draw a third historigram for the net balance on the same graph. A
portion of this graph, (corresponding to the negative values of the net balance) will be below the X-axis.
Method 2. Draw two historigrams, one for each phenomenon (variable), on the same graph. The net
balance between the variables is depicted by proper filling or shading of the space between the two
historigrams, depicting clearly the positive and negative balance.
Both these methods are explained in the following illustration.
4·44
BUSINESS STATISTICS
Example 4·33. India’s overall balance of payment situation (Billions of Rupees) is given below :
Years :
Credits
Debits
Balance (Credit – Debit)
1970—71
18·9
22·2
–3·3
1971—72
20·9
24·9
– 4·0
1972—73
24·2
26·7
– 2·5
1973—74
46·1
33·0
13·1
1974—75
40·7
47·2
–6·5
Represent the above data on the same graph.
Solution. The above data can best be represented by the Silhouette or Net Balance Graph. The graphs
obtained on using both the methods are given a below.
Favourable
Unfavourable
0
20
0
1974-75
10
1973-74
10
Debits
30
1972-73
20
Credits
40
1971-72
30
50
1970-71
40
1974-75
1973-74
1972-73
1971-72
YEARS
1970-71
RUPEES (IN BILLIONS)
50
–10
Method 2
CREDITS AND DEBITS DURING 1970—1975
(ALONG WITH BALANCE OF PAYMENTS)
RUPEES (IN BILLIONS)
Method 1
INDIA’S OVERALL BALANCE OF PAYMENTS
Credits
Debits
Balance
Fig. 4·47.
YEARS
Fig. 4·46.
C. RANGE GRAPH OR ZONE GRAPH
By range we mean the deviation, i.e., difference between the two extreme values viz., the maximum
and minimum values of the variable under consideration.
Range graph, also sometimes known as zone graph, is used to depict and emphasize the range of
variation of a phenomenon for each period. For instance, for highlighting the range of variation of :
(i) the temperature on different days,
(ii) the blood pressure readings of an individual on different days,
(iii) price of a commodity on different periods of time, etc.,
the range or zone graph is the most appropriate and helps us to have an idea of the likely fluctuations in the
magnitudes of the phenomenon under study. The range chart can be drawn in any of the following ways.
Method 1. For each time period, plot the maximum and minimum values of the variable and join them
by straight lines to get the range lines. Plot the mid-point (average value of the variable) for each period.
Join these points by straight lines to get the range graph. The range graph thus depicts the maximum, the
minimum and the average value of the phenomenon for each period.
Method 2. This method consists in plotting two historigrams, one corresponding to the maximum
values of the phenomenon for different periods and the other corresponding to the minimum values. The
4·45
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
space between the two historigrams depicts the range of the variation and is prominently displayed by
proper filling or shading it.
Both these methods are explained in the following illustration.
Example 4·34. The following are the share price quotations of a firm for five consecutive weeks.
Present the data by an appropriate diagram.
Week
High
Low
1
102
100
2
103
101
3
107
103
4
106
105
5
105
104
Solution. Since the maximum and minimum price quotations of a firm for 5 consecutive weeks are
given, the most appropriate graph for it is the zone or the range graph.
Taking weeks along X-axis and share price quotations along Y-axis and using the false base line at 100
for the vertical scale, the range chart as obtained by Method 1 is drawn in Fig. 4·48.
RANGE CHART FOR SHARE PRICE
QUOTATIONS OF A FIRM
107
Highest
SHARE-PRICE QUOTATIONS
SHARE-PRICE QUOTATIONS
Average
104
Lowest
103
Highest Price
Lowest Price
107
106
105
RANGE CHART FOR SHARE PRICE
QUOTATIONS OF A FIRM
106
105
104
103
102
101
100
102
0
101
1
2
3 4
WEEKS
5
Fig. 4·49.
100
0
1
2
3
WEEKS
4
5
Fig. 4·48.
The range chart may also be drawn as in Figure 4·49 : (c.f., Method 2).
D. BAND GRAPH
Like sub-divided bar diagram or pie diagram, the band graph, also known as component part line
chart, is a line graph used to display the total value or the magnitude of a variable and its break up into
different components for each period. The construction of such a chart which is used only for time series is
quite simple and involves the following steps :
(i) For each period, arrange the break up of the value of the variable into various components in the
same order.
(ii) Draw historigram for the first component.
4·46
BUSINESS STATISTICS
(iii) Over this historigram draw another historigram for the 2nd component. This is done by drawing
the 2nd historigram for the cumulative totals of the first two components.
(iv) Over the 2nd historigram draw another historigram for the third component. This is done by
drawing the historigram for the cumulative totals of the first three components. This technique of drawing
historigrams, one over the other is continued till all the components are exhausted. The last historigram,
thus corresponds to the total value of the variable.
The space between different historigrams in the form of different bands or belts, one for each
component, is prominently displayed by different types of lines viz., dash lines, dot lines, dash-dot lines,
etc. This chart is specially useful to display the division of the total costs, total sales, total production, etc.,
into various component parts for different periods.
Remark. Just like percentage bar diagram or percentage rectangular diagram, band chart can also be
used for time series where data are expressed in percentage form. In such a situation, the total value of the
variable for each period is taken as 100 and bands will depict the percentage that different components bear
to the total.
1990—91
1991—92
1992—93
1993—94
1994—95
1995—96
1996—97
1997—98
Material
Labour
Overhead
Total
1989—90
Items
1988—89
Example 4·35. The following table gives the cost of production (in arbitrary units) of a factory in
biennial averages :
37
10
13
60
25
8
10
43
35
11
15
61
36
11
16
63
35
11
17
63
38
12
20
70
22
7
12
41
17
5
9
31
26
8
12
46
20
9
15
44
Represent the above data by a band graph.
Solution.
COST OF PRODUCTION OF A FACTORY
(1988—1998)
70
60
Labour
40
30
20
Materials
YEARS
Fig. 4·50
1997-98
1996-97
1995-96
1994-95
1993-94
1992-93
1991-92
1990-91
0
1989-90
10
1988-89
COST OF PRODUCTION
Overhead
50
4·47
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
4·4·5. Semi-Logarithmic Line Graphs or Ratio Charts. In the graphs discussed so far, we have used
arithmetic i.e., natural scale in which equal distances represent equal absolute magnitudes on both the axes.
Such graphs can be used with advantage if we are interested in displaying the absolute changes in the value
of a phenomenon and the variations in the magnitudes are such that they can be plotted in the available
space in the graph paper. But quite often, particularly in the case of phenomena pertaining to growth like
population, production, sales, profits, etc., the increase or decrease in the value of the variable is very rapid.
In such a situation we are primarily interested to study the relative changes rather than the absolute changes
in the value of the phenomenon and the arithmetic scale is not of much use. In such cases we use semilogarithmic or logarithmic or ratio scale which is basically used to highlight or emphasize relative or
proportionate or percentage changes in the values of a phenomenon over different periods of time.
Since
log
()
a
= log a – log b,
b
on a logarithmic scale, equal distances will represent equal proportionate changes.
There are two ways of using logarithmic scale :
Semi-logarithmic Line Graph. In such a graph, the time variable along X-axis is expressed on a
natural scale and the logarithms of the values of the phenomenon under study for different periods of time
are plotted on the vertical axis on a natural scale. The points so obtained are joined by straight lines to give
the desired curve. Since, in this type of curve the logarithms are taken along only one axis, it is known as
semi-logarithmic graph and it is specially useful for studying the rates of change in the dependent variable
(phenomenon under study) for different periods of time in a time series.
Logarithmic Line Graph. In this graph, both the variables along horizontal and vertical axis are
plotted on a logarithmic scale. For instance, for a time series data, the logarithms of the time values are
plotted along horizontal axis and the logarithms of the values of the variable are plotted along the vertical
axis, each on a natural scale. The required graph is obtained on joining the points so obtained by straight
lines. However, it is very difficult to interpret such a graph and in practice, mostly semi-logarithmic graph
is used.
Remarks 1. In a semi-logarithmic graph, almost always, the vertical scale or Y-scale is a logarithmic
scale. Since a semi-logarithmic graph is useful for studying the relative changes or rates and ratios of
increase or decrease over different periods of time, it is also called a Ratio Graph or Ratio Chart and the
logarithmic scale is also called ratio scale.
2. For practical purposes, semi-logarithmic graph papers (in which vertical scale is logarithmic scale
i.e., log Y is marked along Y-axis and horizontal scale is natural i.e., the values of X are marked in
arithmetical scale), analogous to ordinary graph paper are available. The use of such a semi-logarithmic
graph paper relieves us of the problem of looking up and plotting the logarithms of the values of a variable
on a natural scale. The specimen of a semi-logarithmic graph paper is given below.
SEMI-LOGARITHMIC GRAPH PAPER
100
80
60
40
100
80
60
40
20
20
10
8
6
4
10
8
6
4
2
2
1
1
Fig. 4·51.
4·48
BUSINESS STATISTICS
3. The following diagram (Fig. 4.52) displays the arithmetic and logarithmic scales.
7—
1·00000
·95424 —
·90309 —
·84510 —
·77815 —
·69897 —
10 —
9—
8—
7—
6—
5—
6—
·60206 —
4—
5—
·47712 —
3—
·30103 —
2—
10 —
9—
8—
4—
3—
2—
1—
·00000
Logarithms on
Arithmetic scale
0
Arithmetic
scale
1—
Logarithmic
scale
Fig. 4·52.
The reason why the logarithmic scale shows lower and lower distances as we move towards higher and
higher magnitudes from 1 to 10 is that the series which is increasing by equal absolute amounts (on an
arithmetic scale) is increasing at a diminishing rate.
Arithmetic Scale Graphs Vs. Ratio Scale Graphs
1. A line graph on an arithmetic scale depicts the absolute changes from one period to another whereas
on a ratio scale it reflects the rate of change between any two points of time. Thus the graph drawn on
natural scale will not be able to reflect the relative or percentage changes or the rate of change of the
phenomenon for any two points of time. In most of the problems of growth, e.g., data relating to
population, production, sales or profits of a business concern, national income, etc., absolute changes if
shown on the graph on a natural scale, are often misleading. As an illustration, let us consider the following
hypothetical figures relating to the profits of a business concern.
Year
1990
1991
1992
1993
1994
1995
Profits
(Rs. in lacs)
15
30
50
75
105
140
Increase over profits of preceding year
Absolute (Rs.. in lacs)
Percentage
—
—
15
100·0
20
66·7
25
50·0
30
40·0
35
33·3
Thus in the above table, although the absolute increase shows a steady increase in the profits, the
percentage or relative increase registers a steady decline. It is surprising to note that the smallest percentage
increase (for the year 1995) corresponds to the greatest absolute increase, a fact which is prominently
displayed on a semi-logarithmic graph by the flattening of the slope of the curve. Hence, if the primary
objective is to study the rate of change in the magnitudes of a phenomenon, the data plotted on a natural
scale will give quite wrong and misleading conclusions. In such a case the ratio or semi-logarithmic scale is
the appropriate one.
2. On an arithmetic or natural scale equal absolute amounts (along vertical axis) are represented by
equal distances whereas in a ratio scale equal distances represent equal proportionate movements or equal
4·49
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
relative rates of change or equal percentage changes. Thus in a natural scale, the readings are in arithmetical
progression while in a ratio scale they are in geometric progression as exhibited in the Fig. 4·53.
5—
32 —
320 —
3200 —
4—
16 —
160 —
1600 —
3—
8—
80 —
800 —
2—
4—
40 —
400 —
1—
2—
20 —
200 —
0—
Natural scale
1—
10 —
Ratio scale
100 —
Fig. 4·53.
Thus on a logarithmic scale, the distances between the points on the vertical scale represent the
distances of the logarithms of the numbers and not the distances of the numbers themselves.
3. On a natural or arithmetic scale, the vertical scale must start with zero. Since the logarithm of zero is
minus infinity, i.e., since log (0) = – ∞, in a ratio graph there is no zero base line. Thus, in a ratio graph, the
vertical scale starts with a positive number. Further, since log (1) = 0, the value 1 is placed at a zero
distance from the origin i.e., at the origin itself. Hence, in a ratio chart, the origin along the vertical scale is
at 1.
4. In case the magnitudes of the phenomenon under consideration have a very wide range, i.e., the
values differ widely in magnitudes, then ratio graph is more appropriate than the graph on an arithmetic
scale.
5. In interpreting the graphs drawn on a natural scale, the relative position of the curve on the graph is
very significant. But, in interpreting a ratio graph it is the shape, direction and degree of steepness of the
graph (i.e., straight line or a curve sloping upwards or downwards) that matters and not its position.
Accordingly, on a semi-logarithmic scale, the different graphs can be moved up and down without
changing their meaning (interpretation). Hence ratio graph can be effectively used to graph, for purposes of
comparisons, two or more phenomena (variables) which differ widely in their magnitudes or which are
measured even in different units. For instance, for charting the data relating to the population growth,
agricultural or industrial output, prices, profits, sales, etc., the ratio graph or semi-logarithmic graph is more
appropriate. Such comparison, however, might be misleading on a natural scale.
Uses of Semi-Logarithmic Scale or Ratio Scale. From the above discussion, the uses of ratio or semilogarithmic scale may be summarised as follows :
1. For studying the rates of change (increase or decrease) or the relative or percentage changes in the
values of a phenomenon like population, production, sales, profit, income, etc.
2. For charting two or more phenomena differing very widely in their magnitudes.
3. For charting and comparative study of two or more phenomena measured in different units.
4. When we are interested in proportionate or percentage changes rather than absolute changes.
Limitations of Semi-Logarithmic Scale or Ratio Scale
1. Since log (0) = – ∞ and the logarithm of a negative quantity is not defined, the ratio scale cannot be
used to plot zero or negative values. Accordingly, it cannot be used to represent the ‘Net Balance’ or
‘Balance of Trade’ on the graph.
4·50
BUSINESS STATISTICS
2. Another limitation of the ratio scale is that it cannot be used to study the total magnitude and its
break up into component parts of any given phenomenon.
3. It cannot be used to study absolute variations.
4. Lastly, it is quite difficult for a layman to draw and interpret ratio charts. The interpretation of a
semi-logarithmic graph requires great skill and expertise. This is a great handicap in the mass popularity of
ratio or semi-logarithmic graphs.
Shape of the Curve on Semi-Logarithmic Scale and Natural Scale
1. The values of phenomenon increasing by a constant amount will give a straight line rising upward
on an arithmetic scale while on a semi-logarithmic scale it will give an upward rising curve with its slope
steadily declining (which implies a steady decreasing rate). In other words, it will be a curve concave to the
base. This is so because the values increasing by a constant absolute amount increase at a declining rate.
This is shown in the following diagram.
SERIES INCREASING BY A CONSTANT ABSOLUTE AMOUNT
ARITHMETIC SCALE
LOGARITHMIC SCALE
10
9
8
7
6
5
4
3
2
1
0
1990 1992 1994 1996 1998 1999
9
8
7
6
5
4
3
2
1
1990 1992 1994 1996 1998 1999
Fig. 4·54.
2. A time series increasing at a constant rate will give a curve convex to the base (i.e., a curve rising
upwards towards the right with its slope gradually increasing), on a natural scale. However, on a ratio
scale, it will give an upward rising straight line graph as shown in the following diagram.
SERIES INCREASING AT A CONSTANT RATE
ARITHMETIC SCALE
LOGARITHMIC SCALE
9
8
7
6
5
4
3
2
1
0
1990 1992 1994 1996 1998 1999
10
9
8
7
6
5
4
3
2
1
1990 1992 1994 1996 1998 1999
Fig. 4·55.
3. If the time series values decrease by a constant absolute amount the graphs on the two scales will be
like the mirror images of the graphs in case 1, in the reverse order (as shown in the Fig. 4·64) i.e., on a
natural scale it will give a straight line moving downwards (rapidly declining) and on a ratio scale it will
give a curve falling to the right with its slope increasing.
4·51
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
SERIES DECREASING BY CONSTANT ABSOLUTE AMOUNT
ARITHMETIC SCALE
LOGARITHMIC SCALE
10
9
8
7
6
5
4
3
2
1
0
1990 1992 1994 1996 1998 1999
9
8
7
6
5
4
3
2
1
1990 1992 1994 1996 1998 1999
Fig. 4·56
4. Similarly, for a time series decreasing at a constant rate, the graphs on the two scales will be the
mirror images of the graphs in case 2, in the reverse order (as shown in the following diagram) i.e., on an
arithmetic scale, we shall get a curve moving downwards with a declining slope and on a semi-logarithmic
scale we shall get a straight line moving downwards.
SERIES DECREASING BY CONSTANT RATE
ARITHMETIC SCALE
LOGARITHMIC SCALE
10
9
9
8
8
7
6
7
5
6
4
5
3
4
3
2
2
1
1
0
1990 1992 1994 1996 1998 1999
1990 1992 1994 1996 1998 1999
Fig. 4·57.
Interpretation of Semi-Logarithmic or Ratio Curves
1. If the curve is rising upwards, the rate of growth or increase is positive and a curve falling
downwards indicates a decreasing rate.
2. If the curve is nearly a straight line which is ascending, it represents the series increasing, more or
less, at a constant rate. Similarly, a nearly straight line curve which is descending i.e., moving downwards,
represents a series which is decreasing, more or less, at a uniform rate.
3. If the curve rises (falls) steeply at one point of time than at another, it depicts rapid rate of increase
(decrease) at that point than at the other point.
4. If two curves on the same semi-logarithmic graph are parallel to each other, they represent equal
percentage rates of change for each phenomenon.
5. If one curve is steeper than the other on the same ratio-chart, it implies that the first is changing at a
faster rate than the second.
We now give below, some illustrations of the use of ratio or semi-logarithmic graphs.
Example 4·36. A firm reported that its net worth (in lacs Rs.) in the years 1990-91 to 1994-95 was as
follows :
Year
1990-91
1991-92
1992-93
1993-94
1994-95
Net worth
100
112
120
133
147
4·52
BUSINESS STATISTICS
Plot the above data in the form of a semi-logarithmic graph. Can you say anything about the
approximate rate of growth of its net worth ?
Solution. To plot the above data on a semi-logarithmic scale we plot the logarithms of the dependent
variable, (Net Worth), along the vertical
SEMI-LOG GRAPH
axis on a natural scale. The horizontal axis,
(NET WORTH OF A FIRM)
as usual, will represent time variable on a
2·18
natural scale.
——————————————————————
——————————————————————
——————————————————————
2·00
2·05
2·08
2·12
2·17
2·12
2·09
2·06
94-95
2·00
0
The graph is shown in Fig. 4·66.
93-94
2·03
92-93
100
112
120
133
147
91-92
1990-91
1991-92
1992-93
1993-94
1994-95
2·15
log (Y)
90-91
Net Worth (Y)
(in lacs Rs.)
LOGARITHMS
Year
Comments. Since the graph is
ascending throughout, it reflects the
YEARS
increasing rate of growth of the net worth
Fig. 4·58.
for the entire period. However, since the
graph is steepest for the period 1990-91 to 1991-92, it represents the highest rate of positive growth
(increase) during this period. Then however, there is slight decline in the rate of increase for the period
1991-92 to 1992-93. There is again increase in the rate of growth for the period 1992-93 to 1994-95 over
the period 1991-92 to 1992-93. Further, since the graph for period 1992-93 to 1994-95 is almost a straight
line, it represents a constant rate of increase during this period.
Example 4·37. The following table gives the population of India at intervals of 10 years :
Year
Population
1931
27,88,67,430
1941
31,85,39,060
1951
36,09,50,365
1961
43,90,72,582
1971
54,79,49,809
LOGARITHMS
Plot the data on a graph paper. From your graph determine the decade in which the rate of growth of
population was,
SEMI-LOG GRAPH
(i) the slowest.
(POPULATION OF INDIA AT
(ii) the fastest.
INTERVALS OF 10 YEARS)
3·75
S o l u t i o n . Since we are interested in
determining the rate of growth for different
3·70
decades, the appropriate graph will be obtained
3·65
on plotting the data on a semi-logarithmic or
ratio scale. Logarithms of the population values
3·60
are plotted along the vertical axis on a natural
3·55
scale and time variable (decades) are plotted
along the horizontal axis on a natural scale.
3·50
2789
3185
3610
4391
5479
3·45
3·50
3·56
3·64
3·74
The graph is shown in figure 4·59
0
YEARS
Fig. 4·59.
1971
1931
1941
1951
1961
1971
3·45
3·40
1961
——————————————————————
——————————————————————
——————————————————————
1951
log (Y)
1941
Population (Y)
(in lakhs)
1931
Year
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
4·53
Comments. Since the graph is ascending throughout, it reflects an increasing rate of population growth
throughout the entire period. Further, since the graph has a maximum steep during the period 1961—1971,
the rate of growth is maximum during this decade. Again, since the graph has minimum steep for the period
1941—1951, the rate of growth is minimum during this decade.
4·5. LIMITATIONS OF DIAGRAMS AND GRAPHS
Diagrams and graphs are very powerful and effective visual statistical aids for presenting the set of
numerical data but they have their limitations some of which are outlined below :
(i) Diagrams and graphs help in simplifying the textual and tabulated facts and thus may be regarded as
supplementary to statistical tables. They should not be regarded as substitutes for classification, tabulation
and some other forms of presentation of a set of numerical data under all circumstances and for all
purposes. Julin has very elegantly stated this limitation in the following words :
“Graphic statistic has a role to play of its own ; it is not the servant of numerical statistics but it cannot
pretend, on the other hand, to precede or displace it”.
(ii) They give only general idea of data so as to make it readily intelligible and thus furnish only
limited and approximate information. For detailed and precise information we have to refer back to the
original statistical tables. Accordingly, diagrams and graphs should be used to explain and impress the
significance of statistical facts to the general public who find it difficult to understand and follow the
numerical figures. They are, therefore, appealing to a layman who does not have any statistical background
but not to a statistician because they are not amenable to further mathematical treatment and hence are not
of much use to him from analysis point of view.
(iii) They are subjective in character and therefore, may be interpreted differently by different people.
If the same set of data are presented diagrammatically (graphically) on two different scales, the sizes of the
diagrams (graphs) so obtained might differ widely and thus generally, might create wrong and misleading
impressions on the minds of the people. Hence, they are likely to be mis-used by unscrupulous and
dishonest people to serve their selfish motives during advertisements, publicity, etc. Hence, they should not
be accepted on their face value without proper scrutiny and caution.
(iv) All the diagrams and graphs are not easy to construct. Two and three-dimensional diagrams, and
ratio graphs require more time and great amount of expertise and skill for their construction and
interpretation and are not readily perceptible to non-mathematical person.
(v) In case of large figures (observations), such a presentation fails to reveal small differences in them.
(vi) The choice of a particular diagram or graph to present a given set of data requires great expertise,
skill and intelligence on the part of the statistician or the concerned agency engaged in the work. A wrong
type of diagram/graph may lead to very fallacious and misleading conclusions. In this context C.W. Lowe
writes :
“The important point that must be borne in mind at all times is that the pictorial presentation, chosen
for any situation, must depict the true relationship and point out the proper conclusion. Use of an
inappropriate chart may distort the facts and mislead the reader. Above all, the chart must be honest.”
(vii) Diagrammatic presentation should be used only for comparison of different sets of data which
relate either to the same phenomenon or different phenomena which are capable of measurement in the
same unit. They are not useful, if absolute information is to be represented.
EXERCISE 4·2
1. (a) Discuss the utility and limitations of graphic method of presenting statistical data.
(b) Discuss the advantages and limitations of representing statistical data by diagrams (including graphs).
(c) What are the general rules of graphical presentation of data ?
[C.S. (Foundation), June 2001]
(d) Explain the advantages of graphic representation of statistical data.
2. (a) What are various types of graphs used for presenting a frequency distribution. Discuss briefly their
(i) construction and (ii) relative merits and demerits.
(b) Explain briefly the various methods that are used for graphical representation of frequency distribution.
4·54
BUSINESS STATISTICS
3. Give an illustration each of the type of data for which you would expect the frequency curve to be :
(i) fairly symmetrical,
(ii) positively skewed,
(iii) negatively skewed,
(iv) J-shaped,
(v) U-shaped.
4. Comment on the following :
(a) “The wandering of a line is more powerful in its effect on the mind than a tabulated statement; it shows what is
happening and what is likely to take place just as quickly as the eye is capable of working.” —Boddington
(b) “Graphs are dynamic, dramatic. They may epitomise an epoch, each dot a fact, each slope an event, each curve
a history ; wherever there are data to record, inferences to draw, or facts to tell, graphs furnish the unrivalled means
whose power we are just beginning to realise and apply.” —Hubbard
5. Explain clearly the distinctions between “natural scale” and “semi-logarithmic scale” used in the graphical
presentation of data.
6. (a) What do you mean by a false base line ? Explain its utility in graphic representation of statistical data.
(b) “A false base line of a graph is a wrong base line.” Comment.
(c) What is false base line ? Under what circumstances should it be used ?
7. Describe briefly the construction of histogram and frequency polygon of a frequency distribution and state their
uses.
Prepare a Histogram and a Frequency Polygon from the following data :
Class :
0—6
6—12
12—18
18—24
24—30
30—36
f
:
4
8
15
20
12
6
8. Draw a histogram of the following distribution :
Life of Electric Lamp (in hours) (Mid-values) :
1010
1030
1050
1070
1090
Firm A (No. of lamps)
:
10
130
482
360
18
9. From the following data draw (i) Histogram, (ii) Frequency polygon and (iii) Frequency curve:
Class
0—10
10—20 20—30 30—40 40—50 50—60 60—70 70—80 80—90
Frequency
4
6
7
14
16
14
8
6
5
[C.S. (Foundation), June 2000]
10. Draw a histogram from the following data :
Daily wages (Rs.)
:
10—20
20—30
30—40
40—60
60—80
80—110
No. of employees
:
5
10
12
28
20
24
[C.S. (Foundation), June 2002]
11. Draw a histogram to represent the following distribution :
Monthly income
0— 750
750—1000
1000—1500
1500—2000
2000—3000
No. of families
15
51
199
240
324
Monthly income
3000— 5000
5000 — 7500
7500—10000
10000—15000
15000—25000
No. of families
309
162
66
50
27
Total
1443
How many families can be expected to have monthly income between 3500 and 4250 rupees.
Hint :
4250 – 3500
× 309 = 115·88 ~– 116.
2000
12. (a) What are cumulative frequencies ? How do you present them diagrammatically for discrete and continuous
distributions ?
(b) What is a cumulative frequency curve ? Mention its kinds. Take an example to illustrate them.
(c) Explain the difference between a histogram and frequency polygon. What is an ogive curve ? State the purpose
for which it is used.
13. Draw the ‘less than’ and ‘more than’ ogive curves from the data given below :
Weekly Wages (’00 Rs.) :
0—20
20—40
40—60
60—80
80—100
No. of Workers
:
10
20
40
20
10
[C.S. (Foundation), Dec. 2000]
4·55
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
14. For the following distribution of wages, draw ogive and hence find the value of median.
Monthly Wages
Frequency
Monthly Wages
Frequency
12·5—17·5
2
37·5—42·5
4
17·5—22·5
22
42·5—47·5
6
22·5—27·5
10
47·5—52·5
1
27·5—32·5
14
52·5—57·5
1
———————
32·5—37·5
3
Total
63
———————
Ans. Md. = 26 (approx.)
15. Age distribution of 200 employees of a firm is given below. Construct a less than ogive curve and hence or
otherwise calculate semi-inter-quartile range (Q3 – Q1) / 2 of the distribution.
Age in years (less than)
No. of employees
:
:
25
10
30
25
35
75
40
130
45
170
50
189
55
200
16. The following table gives the distribution of the wages of 65 employees in a factory :
Wages in Rs.
(Equal to or more than)
50
60
70
80
90
100
110
120
Number of Employees
65
57
47
31
17
7
2
0
Draw a ‘less than’ ogive curve from the above data, and estimate the number of employees earning at least Rs. 63
but less than Rs. 75.
Ans. 15
17. Draw a less than cumulative frequency curve of the following distribution and find the limits for the central
60% of the distribution from the graph.
x (less than)
:
5
10
15
20
25
30
35
40
45
Frequency
:
2
11
29
45
69
83
90
96
100
Ans. 13 to 29·8.
18. Draw a less than ogive from the following data :
Weekly Income (Rs.)
12,000 11,000 10,000
8,000
6,000
4,000
3,000
2,000
1,000
(equal to or more than)
No. of Families
0
6
14
26
42
54
62
70
80
From the graph estimate the number of families in the income range of Rs. 2,400 and Rs. 10,500. Also find
maximum income of the lowest 25% of the families.
[Delhi Univ., B.Com,. (Hons.), 2006]
Ans. Income (in Rs. ’000) 10—20 20—30 30—40 40—60 60–80 80—100 100—110 110—120
No. of families
10
8
8
12
16
12
8
6
57 (Approximately); Q1 = Rs. 3,250.
19. The monthly profits (Rs. lakhs) earned by 100 companies during the financial year 2002-03 are given in the
table below :
Monthly Profit (Rs. lakhs) : 20—30
30–40
40–50
50—60 60—70 70—80 80—90 90—100
Number of companies
:
4
8
18
30
15
10
8
7
Draw the OGIVE by “less than method” and “more than method.”
[Delhi Univ., (FMS), M.B.A., March 2004]
20. Construct a frequency table for the following data regarding annual profits, in thousands of rupees in 50 firms,
taking 25—34, 35—44, etc., as class intervals.
28
35
61
29
36
48
57
67
69
50
48
40
47
42
41
37
51
62
63
33
31
32
35
40
38
37
60
51
54
56
37
46
42
38
61
59
58
44
39
57
38
44
45
45
47
38
44
47
47
64
Construct a less than ogive and find :
(i) Number of firms having profit between Rs. 37,000 and Rs. 58,000.
(ii) Profit above which 10% of the firms will have their profits.
(iii) Middle 50% profit group.
Ans. (i) 30, (ii) Rs. 62,000, (iii) Rs. 39,000 to Rs. 56,000.
4·56
BUSINESS STATISTICS
21. What do you mean by a historigram ? How does it differ from histogram ?
22. (a) What are different types of graphs commonly used to present a time series data ? Bring out their salient
features.
(b) Describe briefly :
(i) Historigram,
(ii) Silhouette or Net Balance Graph,
(iii) Range Graph,
(iv) Band Graph,
for presenting time series data.
23. Represent the data relating to consolidated budgetary position of states in India as given below, on a graph
paper.
(Rs. crore)
Year
Revenue
Expenditure
Surplus or Deficit
1955-56
560·1
626·4
–66·3
1956-57
577·0
654·3
–77·3
1957-58
705·6
677·3
+28·3
1958-59
742·1
745·8
–3·7
1959-60
833·9
829·9
+4·0
Also depict graphically, the net balance of trade.
24. Represent the following data by means of a time series graph.
Year
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
Export (Rs. ’000)
267
269
263
275
270
280
282
272
265
266
Import (Rs. ’000)
307
310
280
260
275
271
280
280
260
265
Show also the net balance of trade.
25. (a) What is a false base line? When is it used on an arithmetic line graph ? [Delhi Univ. B.Com. (Pass), 2001]
(b) Prepare a graph of the following data by using a false base line.
Centre
All India
Delhi
1969
177
185
Consumer Price Index Numbers (1960 = 100)
Years
1970
1971
1972
1973
186
192
207
250
199
211
222
265
1974
360
337
26. Present the following hypothetical data graphically.
AREA AND PRODUCTION OF RICE IN INDIA
Year
:
1987
1988
1989
1990
1991
1992
Area (Million Acres)
:
174·1
177·3
176·1
177·9
179·3
179·1
Production (Million Tonnes)
:
72·5
77·8
74·8
77·2
78·0
74·8
27. Marks obtained by 100 students in Economics are given below. Draw an appropriate graph to represent them :
Marks
Males
Females
Total
30—40
8
6
14
40—50
6
10
16
50—60
14
6
20
60—70
13
12
25
70—80
12
13
25
Total
53
47
100
[C.S. (Foundation). Dec. 1999]
Hint: Draw three graphs for males, females and total on the same graph paper.
28. Present the following data about India by a suitable graph :
PRODUCTION IN MILLION TONS
Year
Rice
Wheat
Pulses
Other Cereals
Total
1962
30
10
10
14
64
1963
32
11
8
18
69
1964
33
8·5
11·5
20
73
1965
35
12
11
20
78
1966
36
10
10
22
78
1967
38
11
9
23
81
Hint: Band Graph
———————
———————
———————
———————
———————
———————
4·57
DIAGRAMMATIC AND GRAPHIC REPRESENTATION
29. Present the following data by a suitable graph :
MINIMUM AND MAXIMUM PRICE OF GOLD FOR 10 GMS. FOR THE YEAR 1967
Months
Highest Price
Lowest Price
Months
Highest Price
Lowest Price
(Rs.)
(Rs.)
(Rs.)
(Rs.)
January
160·0
152·0
July
175·0
163·2
February
162·2
156·0
August
175·8
160·0
March
165·0
160·3
September
172·2
165·0
April
166·5
162·4
October
178·0
168·0
May
168·2
160·5
November
171·0
165·0
June
170·0
161·9
December
175·5
167·0
Hint: Range Graph.
30. (a) Differentiate between the natural scale and logarithmic scale used in graphic presentation of data. In which
cases should the latter scale be used ?
(b) Explain what is meant by semi-logarithmic diagram and discuss its advantages over the natural scale diagram.
(c) Explain briefly how you will interpret the graphs drawn on a semi-logarithmic scale.
(d) What do you understand by a ratio-scale ? Under what situations ratio charts should be drawn ?
31. The following table shows the total sales of Gold Bonds by the Reserve Bank of India :
Month
Year
Rs. (’000)
Month
Year
October
1965
15,560
April
1966
November
1965
13,170
May
1966
December
1965
18,740
June
1966
January
1966
12,450
July
1966
February
1966
8,320
August
1966
March
1966
7,540
September
1966
Rs. (’000)
3,250
3,570
3,620
3,140
2,580
2,540
Represent the data graphically on the logarithmic scale
32. Plot the following data graphically on the logarithmic scale.
Year
Total notes issued
(in crores Rs.)
2890
3065
3242
3536
3866
1965—66
1966—67
1967—68
1968—69
1969—70
Total notes in circulation
(in crores Rs.)
2866
3020
3194
3497
3843
33. Present the following data graphically and comment on the features thus revealed :
Year
Production of steel plates
(in thousand tons)
Unit A
30
1990
29
1992
31
1994
30
1996
30
1998
30
2000
30
2002
How will the graph look like if the data are plotted on semi-logarithmic scale ?
Unit B
40
20
10
20
30
40
60
5
Averages or Measures of
Central Tendency
5·1. INTRODUCTION
One of the important objectives of statistical analysis is to determine various numerical measures
which describe the inherent characteristics of a frequency distribution. The first of such measures is
average. The averages are the measures which condense a huge unwieldy set of numerical data into single
numerical values which are representative of the entire distribution. In the words of Prof. R.A. Fisher, “The
inherent inability of the human mind to grasp in its entirety a large body of numerical data compels us to
seek relatively few constants that will adequately describe the data”. Averages are one of such few
constants. Averages provide us the gist and give a bird’s eye view of the huge mass of unwieldy numerical
data.
Averages are the typical values around which other items of the distribution congregate. They are the
values which lie between the two extreme observations, (i.e., the smallest and the largest observations), of
the distribution and give us an idea about the concentration of the values in the central part of the
distribution. Accordingly they are also sometimes referred to as the Measures of Central Tendency.
Averages are very much useful :
(i) For describing the distribution in concise manner.
(ii) For comparative study of different distributions.
(iii) For computing various other statistical measures such as dispersion, skewness, kurtosis and
various other basic characteristics of a mass of data.
Remark. Averages are also sometimes referred to as Measures of Location since they enable us to
locate the position or place of the distribution in question.
We give below some definitions of an average as given by different statisticians from time to time.
WHAT THEY SAY ABOUT AVERAGES — SOME DEFINITIONS
“Averages are statistical constants which enable us to comprehend in a single effort the
significance of the whole.”—A.L. Bowley
“An average is a single value selected from a group of values to represent them in some
way, a value which is supposed to stand for whole group of which it is part, as typical of all the
values in the group.”—A.E. Waugh
“An average value is a single value within the range of the data that is used to represent all
of the values in the series. Since an average is somewhere within the range of the data, it is
sometimes called a measure of central value.”—Croxton and Cowden
“An average is sometimes called a measure of central tendency because individual values of
the variable usually cluster around it. Averages are useful, however, for certain types of data in
which there is little or no central tendency.”—Crum and Smith
“Statistical analysis seeks to develop concise summary figures which describe a large body
of quantitative data. One of the most widely used set of summary figures is known as measures
of location, which are often referred to as averages, measures of central tendency or central
location. The purpose for computing an average value for a set of observations is to obtain a
single value which is representative of all the items and which the mind can grasp simply and
quickly. The single value is the point or location around which the individual items cluster.”
—Lawrence J. Kaplan
5·2
BUSINESS STATISTICS
5·2. REQUISITES OF A GOOD AVERAGE OR MEASURE OF CENTRAL
TENDENCY
According to Prof. Yule, the following are the desideratta (requirements) to be satisfied by an ideal
average or measure of central tendency :
(i) It should be rigidly defined i.e., the definition should be clear and unambiguous so that it leads to
one and only one interpretation by different persons. In other words, the definition should not leave
anything to the discretion of the investigator or the observer. If it is not rigidly defined then the bias
introduced by the investigator will make its value unstable and render it unrepresentative of the distribution.
(ii) It should be easy to understand and calculate even for a non-mathematical person. In other words,
it should be readily comprehensible and should be computed with sufficient ease and rapidity and should
not involve heavy arithmetical calculations. However, this should not be accomplished at the expense of
accuracy or some other advantages which an average may possess.
(iii) It should be based on all the observations. Thus, in the computation of an ideal average the entire
set of data at our disposal should be used and there should not be any loss of information resulting from not
using the available data. Obviously, if the whole data is not used in computing the average, it will be
unrepresentative of the distribution.
(iv) It should be suitable for further mathematical treatment. In other words, the average should
possess some important and interesting mathematical properties so that its use in further statistical theory is
enhanced. For example, if we are given the averages and sizes (frequencies) of a number of different groups
then for an ideal average we should be in a position to compute the average of the combined group. If an
average is not amenable to further algebraic manipulation, then obviously its use will be very much limited
for further applications in statistical theory.
(v) It should be affected as little as possible by fluctuations of sampling. By this we mean that if we
take independent random samples of the same size from a given population and compute the average for
each of these samples then, for an ideal average, the values so obtained from different samples should not
vary much from one another. The difference in the values of the average for different samples is attributed
to the so-called fluctuations of sampling. This property is also explained by saying that an ideal average
should possess sampling stability.
(vi) It should not be affected much by extreme observations. By extreme observations we mean very
small or very large observations. Thus a few very small or very large observations should not unduly affect
the value of a good average.
5·3. VARIOUS MEASURES OF CENTRAL TENDENCY
The following are the five measures of central tendency or measures of location which are commonly
used in practice.
(i) Arithmetic Mean or simply Mean
(ii) Median
(iii) Mode
(iv) Geometric Mean
(v) Harmonic Mean
In the following sections we shall discuss them in detail one by one.
5·4. ARITHMETIC MEAN
Arithmetic mean of a given set of observations is their sum divided by the number of observations. For
example, the arithmetic mean of 5, 8, 10, 15, 24 and 28 is
5 + 8 + 10 + 15 + 24 + 28
90
= 6
6
= 15
In general, if X1, X2,…, Xn are the given n observations, then their arithmetic mean, usually denoted by
X is given by :
– X1 + X2 + … + Xn ∑X
X=
= n
…(5·1)
n
–
5·3
AVERAGES OR MEASURES OF CENTRAL TENDENCY
where ∑X is the sum of the observations.
In case of frequency distribution :
X1 X2
X
f
f1
f2
X2
…
Xn
——————————————————————————————————
f3
…
fn
–
the arithmetic mean X is given by :
(X1 + X1 + … f1 times) + (X 2 + X2 + … f2 times) + . + (Xn + Xn + … fn times.)
–
X =
f1 + f2 + … + fn
f1 X1 + f2X2 + … + fn Xn ∑fX
∑fX
=
=
=
…(5·2)
f1 + f2 + … + fn
∑f
N
where N = ∑f, is the total frequency.
In case of continuous or grouped frequency distribution, the value of X is taken as the mid-value of the
corresponding class.
Remark. The symbol ∑ is the letter capital sigma of the Greek alphabet and is used in mathematics to
denote the sum of values.
Steps for the Computation of Arithmetic Mean
1. Multiply each value of X or the mid-value of the class (in case of grouped or continuous frequency
distribution) by the corresponding frequency f.
2. Obtain the total of the products obtained in step 1 above to get ∑fX.
3. Divide the total obtained in step 2 by N = ∑f, the total frequency.
The resulting value gives the arithmetic mean.
Example 5·1. The intelligence quotients (IQ’s) of 10 boys in a class are given below :
70, 120, 110, 101, 88, 83, 95, 98, 107, 100
Find the mean I.Q.
–
Solution. Mean I.Q. (X) of the 10 boys is given by :
– ∑X 1
X = n = 10 (70 + 120 + 110 + 101 + 88 + 83 + 95 + 98 + 107 + 100) = 972
10 = 97·2
Example 5·2. The following is the frequency distribution of the number of telephone calls received in
245 successive one-minute intervals at an exchange :
Number of Calls
:
0
1
2
Frequency
:
14
21
25
Obtain the mean number of calls per minute.
3
43
4
51
5
40
6
39
7
12
Solution. Let the variable X denote the number of calls received per minute at the exchange.
COMPUTATION OF MEAN NUMBER OF CALLS
No. of Calls (X)
Frequency (f)
fX
0
14
0
1
21
21
2
25
50
3
43
129
4
51
204
5
40
200
6
39
234
7
12
84
Total
N = 245
∑fX = 922
Mean number of calls per minute at the exchange is given by :
– ∑fX 922
X = N = 245 = 3·763
5·4·1. Step Deviation Method for Computing Arithmetic Mean. It may be pointed out that the
formula (5·2) can be used conveniently if the values of X or/and f are small. However, if the values of X
or/and f are large, the calculation of mean by the formula (5·2) is quite tedious and time consuming. In such
5·4
BUSINESS STATISTICS
a case the calculations can be reduced to a great extent by using the step deviation method which consists in
taking the deviations (differences) of the given observations from any arbitrary value A.
Let
d =X–A
…(5·3)
–
∑fd
then,
X =A+
…(5·4)
N
This formula is much more convenient to use for numerical problems than the formula (5·2).
In case of grouped or continuous frequency distribution, with class intervals of equal magnitude, the
calculations are further simplified by taking :
X–A
d = h
…(5·5)
where X is the mid-value of the class and h is the common magnitude of the class intervals. Then
–
∑fd
X =A+h
…(5·6)
N
Steps for Computation of Mean by Step Deviation Method in (5·6)
Step 1. Compute d = (X – A)/h, A being any arbitrary number and h is the common magnitude of the
classes. Algebraic signs + or – are to be taken with the deviations.
Step 2. Multiply d by the corresponding frequency f to get fd.
Step 3. Find the sum of the products obtained in step 2 to get ∑fd.
Step 4. Divide the sum obtained in step 3 by N, the total frequency.
Step 5. Multiply the value obtained in step 4 by h.
Step 6. Add A to the value obtained in step (5).
The resulting value gives the arithmetic mean of the given distribution.
Remarks 1. If we take h = 1, then formula (5·6) reduces to formula (5·4).
2. Any number can serve the purpose of the arbitrary constant ‘A’ used in (5·4) and (5·6) but generally
the value of X corresponding to the middle part of the distribution will be more convenient. In fact, ‘A’ need
not necessarily be one of the values of X.
Example 5·3. Calculate the mean for the following frequency distribution :
Marks
:
Number of students :
0—10
6
10—20
5
(i) By the direct formula.
Solution.
Marks
0—10
10—20
20—30
30—40
40—50
50—60
60—70
;
20—30
8
30—40
15
40—50
7
∑ fX 1670
= 50 = 33·4 marks.
∑f
(ii) Step Deviation Method : In the usual notations we have A = 35 and h = 10.
h∑fd
–
∴
X = A + N = 35 + 10 ×50(– 8) = 35 – 1·6 = 33·4 marks.
–
Mean (X) =
60—70
3
(ii) By the step deviation method.
COMPUTATION OF ARITHMETIC MEAN
Number of
X – 35
fX
d = 10
Mid-value (X)
Students (f)
–3
30
6
5
–2
75
5
15
–1
200
8
25
0
525
15
35
1
315
7
45
2
330
6
55
3
195
3
65
N = ∑f = 50
∑fX = 1670
(i) Direct Formula :
50—60
6
fd
–18
–10
–8
0
7
12
9
∑fd = – 8
5·5
AVERAGES OR MEASURES OF CENTRAL TENDENCY
5·4·2. Mathematical Properties of Arithmetic Mean. Arithmetic mean possesses some very
interesting and important mathematical properties as given below :
Property 1. The algebraic sum of the deviations of the given set of observations from their arithmetic
mean is zero.
–
Mathematically,
∑(X – X ) = 0,
…(5·7)
–
or for a frequency distribution :
∑f(X – X) = 0
…(5·7a)
—
–
–
Proof. ∑f(X – X ) = ∑(fX – fX) =∑fX – ∑fX
–
–
–
= ∑fX – X ∑ f = ∑fX – X.N
[·. · X is a constant and ∑f = N]
–
– –
– 1
–
∴
∑f(X – X) = NX – X.N = 0
[·. · X = N
∑fX ⇒ ∑fX = NX]
Remarks 1. In computing algebraic sum of deviations, we take into consideration the plus and minus
–
sign of the deviations (X – X) as against the absolute deviations (c.f. Mean Deviation in Chapter 6) where
we ignore the signs of the deviations.
2. Verification of Property 1 from the data of Example 5·3.
ALGEBRAIC SUM OF DEVIATIONS FROM MEAN
–
X
f
X – X = X – 33·4
–28·4
6
5
–18·4
5
15
– 8·4
8
25
1·6
15
35
11·6
7
45
21·6
6
55
31·6
3
65
Marks
0—10
10—20
20—30
30—40
40—50
50—60
60—70
–
f (X – X)
–170·4
– 92·0
– 67·2
24·0
81·2
129·6
94·8
–
∑f(X – X) = 0
–
Thus
∑f(X – X) = 0, as required.
It should be kept in mind that in case of the values of the variable when no frequencies are given, we
–
will get ∑(X – X) = 0.
As a simple illustration, let us consider the following case.
X
1
2
3
4
5
6
7
–
X– X = X – 4
–3
–2
–1
0
1
2
3
∑ X = 28
–
∑(X – X ) = 0
– ∑X 28
X = n = =4
–
7
Hence
∑(X – X ) = 0.
Property 2. Mean of the Combined Series. If we know the sizes and means of two component series,
then we can find the mean of the resultant series obtained on combining the given series.
– –
–
If n1 and n2 are the sizes and X1 , X2 are the respective means of two groups then the mean X of the
combined group of size n 1 + n2 is given by :
–
–
–
n1X1 + n2X2
X =
…(5·8)
n1 + n 2
–
Proof. We know that if X is the mean of n observations
–
–
then
X = ∑X
⇒
∑X = nX
n
i.e., Sum of n observations = n × Arithmetic Mean
…(*)
5·6
BUSINESS STATISTICS
–
–
If X1 is the mean of n 1 observations of the first group and X2 is the mean of n 2 observations of the
second group, then on using (*), we get
–
–
The sum of n1 observations of the first group = n1 × X1 = n1X1
–
–
The sum of other n2 observations of the second group = n2 × X2 = n2X2
–
–
∴ The sum of (n1 + n 2) observations of the combined group = n1X 1 + n2X 2
…(**)
–
Hence, the mean X of the combined group n1 + n2 observations is given by :
–
–
Sum of (n1 + n2 ) observations n1 X1 + n2X2
–
X =
=
[From (**)]
n1 + n2
n1 + n2
–
Remarks 1. Some writers use the notation X12 for the combined mean of two groups and thus we may
write :
–
–
–
n X + n2X2
X12 = 1 1
…(5·8a)
n1 + n2
– –
–
2. Generalisation. In general, if X 1 , X 2 ,…,X k are the arithmetic means of k groups with n1 , n2 ,…,nk
–
observations respectively then we can similarly prove that the mean X of the combined group of size
n1 + n2 + … + nk is given by :
–
–
–
n1 X1 + n2X2 + …+ nkXk
–
X =
…(5·8b)
n1 + n2 + … + n k
Property 3. The sum of the squares of deviations of the given set of observations is minimum when
taken from the arithmetic mean.
Mathematically, for a given frequency distribution, the sum
S = ∑f(X – A)2,
… (5·9)
which represents the–sum of the squares of deviations of given observations from any arbitrary value ‘A’ is
minimum when A = X .
This means that, if for any set of data, we compute :
–
S1 = Sum of squared deviations from mean = ∑ (X – X )2 , and
S = Sum of squared deviations from any arbitrary point A
–
= ∑ (X – A) 2 ; A ≠ X ,
then S1 is always less than S i.e., S1 < S.
Illustration of Property 3. Let us consider the values of the variable X as 1, 2, 3, 4, 5, 6, 7.
TABLE 1. SUM OF SQUARED DEVIATIONS
FROM MEAN
–
–
X
1
2
3
4
5
6
7
X – X= X – 4
–3
–2
–1
0
1
2
3
(X – X )2
9
4
1
0
1
4
9
Total 28
∑(X – X ) = 0
–
∑(X – X )2 = 28
–
TABLE 2. SUM OF SQUARED DEVIATIONS
ABOUT ARBITRARY POINT A = 5
X
1
2
3
4
5
6
7
Total
X–A=X–5
–4
–3
–2
–1
0
1
2
(X – A)2
16
9
4
1
0
1
4
∑(X – A)2 = 35
5·7
AVERAGES OR MEASURES OF CENTRAL TENDENCY
28
Mean (X ) = ∑X
7 = 7 =4
–
The sum of the squared deviations of given observations from their mean, in this case is
–
∑(X – X)2 = 28. (From Table 1)
–
For the above case we take the deviations of the values X from any arbitrary point A, (A ≠ X) and then
–
compute the sum of squared deviations about A viz., ∑(X – A)2, A ≠ X; then this sum will be greater than 28
–
for all values of A. Let us in particular take A = 5, (not equal to mean X = 4).
∑(X – A)2 = ∑(X – 5 )2 = 35,
Thus
(From Table 2)
which is greater than the sum of squared deviations about mean viz., 28.
– ∑ fX
–
Property 4. We have :
X= N
⇒
∑ fX = NX
…(5·10)
Result (5·10) is quite useful in the following problems :
–
(a) If we are given the mean wages (X) of a number of workers (N) in a factory, then using (5·10) we
can determine the total wage bill of the factory.
–
(b) Wrong Observations. Suppose we compute the mean X of N observations and later on it is found
that one, two or more of the observations were wrongly copied down. It is now required to compute the
corrected mean by replacing the wrong observations by the correct
– ones. By using (5·10), we can obtain the
uncorrected sum of the observations which is given by N X. From this, if we subtract the wrong
observations, say, X1′ and X2 ′ and add the corresponding correct observations, say, X1 and X2 we can obtain
the corrected sum of the observations which will be given by
–
NX – (X1′ + X2 ′) + X 1 + X2
Dividing this by N, we get the corrected mean.
In general, if r observations are misread as X1′, X2 ′,…, Xr′ while correct observations are X1 , X2,…,Xr ,
then the corrected sum of observations is given by
–
NX – (X1′ + X2 ′ + … + Xr′) + (X1 + X2 + … + Xr)
Dividing this sum by N, we get the corrected mean.
For numerical illustration see Example 5·11.
Remark. If all the observations of a series are added, subtracted, multiplied or divided by a constant
β, the mean is also added, subtracted, multiplied or divided by the same constant.
Let the given observations of the series be x , x ,…, x with mean –x, given by
1
(
2
n
–x = 1 x + x + … + x
1
2
n
n
)
…(*)
Let the new series obtained on adding, subtracting, multiplying and dividing each observation of the
given x-series by a constant β be denoted by the variables U, V, W and Z respectively so that :
U=X+β,
V=X–β,
W = βX ,
–
U = –x + β,
–
–
W = β . –x ,
Z = Xβ
Then
V = –x – β,
–
and Z =
l –
·x
β
Illustration. Let the mean of 10 observations be 35. If each observation is increased by 5, then the new
mean is also increased by 5, i.e., it becomes 35 + 5 = 40. Similarly, if each observation is decreased by 5,
the new mean will be 35 – 5 = 30. Further, if each observation is multiplied by 2, the new mean will be
2 × 35 = 70.
5·8
BUSINESS STATISTICS
5·4·3. Merits and Demerits of Arithmetic Mean
Merits. In the light of the properties laid down by Prof. Yule for an ideal measure of central tendency,
arithmetic mean possesses the following merits :
(i) It is rigidly defined.
(ii) It is easy to calculate and understand
(iii) It is based on all the observations.
(iv) It is suitable for further mathematical treatment. The mean of the combined series is given by (5·8)
or (5·8a). Moreover, it possesses many important mathematical properties (Properties 1 to 4 as discussed
earlier) because of which it has very wide applications in statistical theory.
(v) Of all the averages, arithmetic mean is affected least by fluctuations of sampling. This property is
explained by saying that arithmetic mean is a stable average.
Demerits. (i) The strongest drawback of arithmetic mean is that it is very much affected by extreme
observations. Two or three very large values of the variable may unduly affect the value of the arithmetic
mean. Let us consider an industrial complex which houses the workers and some big officials like general
manager, chief engineer, architect etc. The average salary of the workers (skilled and unskilled) is, say,
Rs. 8,000 per month. If the salaries of the few big bosses (who draw very high salaries) are also included,
the average wage per worker comes out to be Rs. 12,000 say. Thus, if we say that the average salary of the
workers in the factory is Rs. 12,000 p.m. it gives a very good impression and one is tempted to think that
the workers are well paid and their standard of living is good. But the real picture is entirely different. Thus,
in the case of extreme observations, the arithmetic mean gives a distorted picture and is no longer
representative of the distribution and quite often leads to very misleading conclusions. Thus, while dealing
with extreme observations, arithmetic mean should be used with caution.
(ii) Arithmetic mean cannot be used in the case of open end classes such as less than 10, more than 70,
etc., since for such classes we cannot determine the mid-value X of the class intervals unless (i) we estimate
the end intervals or (ii) we are given the total value of the variable in the open end classes. In such cases
mode or median (discussed latter) may be used.
(iii) It cannot be determined by inspection nor can it be located graphically.
(iv) Arithmetic mean cannot be used if we are dealing with qualitative characteristics which cannot be
measured quantitatively such as intelligence, honesty, beauty, etc. In such cases median (discussed later) is
the only average to be used.
(v) Arithmetic mean cannot be obtained if a single observation is missing or lost or is illegible unless
we drop it out and compute the arithmetic mean of the remaining values.
(vi) In extremely asymmetrical (skewed) distribution, usually arithmetic mean is not representative of
the distribution and hence is not a suitable measure of location.
(vii) Arithmetic mean may lead to wrong conclusions if the details of the data from which it is obtained
are not available. In this connection it is worthwhile to quote the words of H. Secrist :
“If an average is taken as a substitute for the details, then the arithmetic mean, in spite of the simplicity
and ease of calculation, has little to recommand when series are non-homogeneous.”
The following example will illustrate this view point.
Let us consider the following marks obtained by two students A and B in three tests, viz., terminal test,
half-yearly examination and annual examination respectively.
Marks in :
Student A
Student B
I Test
55%
65%
II Test
60%
60%
III Test
65%
55%
Average marks
60%
60%
5·9
AVERAGES OR MEASURES OF CENTRAL TENDENCY
The average marks obtained by each of the two students at the end of year are 60%. If we are given the
average marks alone we conclude that the level of intelligence of both the students at the end of the year is
same. This is a fallacious conclusion since we find from the data that student A has improved consistently
while student B has deteriorated consistently.
(viii) Arithmetic mean may not be one of the values which the variable actually takes and is termed as
a fictitious average. Sometimes, it may give meaningless results. In this context it is interesting to quote the
remarks of the ‘Punch’ journal :
“The figure of 2·2 children per adult female was felt to be in some respects absurd, and a Royal
Commission suggested that middle classes be paid money to increase the average to a sounder and more
convenient number”.
Example 5·4. The numbers 3·2, 5·8, 7·9 and 4·5, have frequencies x, (x + 2), (x – 3 ) and (x + 6)
respectively. If the arithmetic mean is 4·876, find the value of x.
Solution.
COMPUTATION OF MEAN
we have :
∑ f = x + (x + 2) + (x – 3) + (x + 6) = 4x + 5
∑ f X = 3·2x + 5·8(x + 2) + 7·9(x – 3) + 4·5(x + 6)
= (3·2 + 5·8 + 7·9 + 4·5)x + 11·6 – 23·7 + 27·0
= 21·4x + 14·9
Number (X)
Frequency (f)
fX
—————————————————————————————————————————————————————————————————————
3·2
5·8
7·9
4·5
x
x+2
x–3
x+6
3·2x
5·8 (x + 2)
7·9 (x – 3)
4·5 (x + 6)
21·4x + 14·9
Mean= ∑∑fX
= 4·876 (Given)
f =
4x + 5
∴
⇒
⇒
⇒
21·4x + 14·9 = 4·876 (4x + 5)
21·4x + 14·9 = 19·504x = 24·380
(21·400 – 19·504)x = 24·380 – 14·900
⇒
1·896x = 9·480
⇒
9·480
x = 1·896 = 5
Example 5·5. In the following grouped data, X are the mid-values of the class intervals and c is a
constant. If the arithmetics mean of the original distribution is 35·84, find its class intervals.
X–c : –21
–14
–7
0
7
14
21
Total
f
:
2
12
19
29
20
13
5
100
[Delhi Univ. B.Com. (Hons.) External), 2007]
Solution. Here X – c is the deviation d from arbitrary point c i.e., d = X – c. Hence, the mean of the
distribution is given by :
∑ fd
∑ f(X – c)
–
X=c+
=
c
+
…(*)
N
N
where N = ∑f.
COMPUTATION OF MEAN AND CLASS INTERVALS
(X – c)
–21
–14
–7
0
7
14
21
f
2
12
19
29
20
13
5
f(X – c)
– 42
–168
–133
0
140
182
105
Total
N = 100
∑ f (X – c) = 84
X
14
21
28
35
42
49
56
Class interval
10·5—17·5
17·5—24·5
24·5—31·5
31·5—38·5
38·5—45·5
45·5—52·5
52·5—59·5
5·10
BUSINESS STATISTICS
Using (*) we get
–
– c)
X = c + ∑ f (X
= 35·84 (Given)
N
∴
84
c + 100
= 35·84
⇒
⇒
X–c=0
⇒
c = 35·84 – 0·84 = 35
X = c = 35
Thus the mid-value of the class corresponding to the value X – c = 0 is X = 35. Further, since the
magnitude of the class interval is 7, the corresponding class interval is obtained on adding and subtracting
(7/2) = 3·5 from 35 and is given by (35 – 3·5, 35 + 3·5) i.e., 31·5—38·5. The class intervals are given in the
last column of the above table.
Example 5·6. Find the class intervals if the arithmetic mean of the following distribution is 33 and
assumed mean 35 :
Step deviation
Frequency
–3
5
–2
10
–1
25
0
30
+1
20
+2
10
Solution. Here the given step deviation is the deviation d, where d = (X – A)/h.
–
X = A + h∑fd
N ; N = ∑ f,
Hence
A = 35 (given)
…(*)
COMPUTATION OF CLASS INTERVALS
Step deviation (d)
–3
–2
–1
0
1
2
Frequency (f)
fd
5
10
25
30
20
10
Total
Using (*),
X
–15
–20
–25
0
20
20
Class Interval
5
15
25
35
45
55
0—10
10—20
20—30
30—40
40—50
50—60
∑fd = –20
N = 100
–
h∑fd
X = A + N = 33 (Given)
20h
33 = 35 – 100
= 35 – 0·2h
⇒
⇒
0·2h = 2
Also
d
= (X h– A)
⇒
h = 10
⇒
X = A + hd = 35 + 10d
Hence, we can calculate different values of X corresponding to the given values of d. Further, since the
magnitude of each class interval is h = 10, we obtain the required C.I.’s by adding 10
2 = 5, to and
subtracting 5 from each mid-value (X), as shown in the last column of the above table.
Example 5·7. From the following data of income distribution calculate the arithmetic mean. It is given
that (i) the total income of persons in the highest group is Rs. 435, and (ii) none is earning less than Rs. 20.
Income (Rs.)
Below 30
” 40
”
50
”
60
No. of persons
16
36
61
76
Income (Rs.)
Below 70
80
”
80
and over
No. of persons
87
95
5
Solution. The open class “Income below 30” includes the persons with income less than Rs. 30. But
since we are given that none is earning less than Rs. 20, this class will be 20—30. Moreover, we are given
the cumulative frequency distribution which has to be converted into the ordinary frequency distribution as
given in the following table :
5·11
AVERAGES OR MEASURES OF CENTRAL TENDENCY
COMPUTATION OF ARITHMETIC MEAN
Income
Mid-value (X)
No. of persons (f)
fX
20—30
30—40
40–50
50—60
60—70
70—80
80 and over
25
35
45
55
65
75
—
16
36 – 16 = 20
61 – 36 = 25
76 – 61 = 15
87 – 76 = 11
95 – 87 = 8
5
400
700
1125
825
715
600
435 *
∑ f = 100
*It
∑ fX = 4,800
is given that total income in the highest group is Rs. 435.
4‚800
Arithmetic Mean = ∑∑fX
f = 100 = Rs. 48.
∴
Example 5·8. An investor buys Rs. 1,200 worth of shares in a company each month. During the first 5
months he bought the shares at a price of Rs. 10, Rs. 12, Rs. 15, Rs. 20 and Rs. 24 per share. After 5 months
what is the average price paid for the shares by him ?
Month
Price per
share (X)
Total Cost
(fX Rs.)
No. of shares bought
(f)
————————————————————
—————————————————————
——————————————————————————
————————————————————————————————
Solution. Let X denote the price
(in Rupees) of a share. Then the
distribution of shares purchased during
the first five months is as follows :
Hence, the average price paid per
share for the first five months is
– ∑ f X 6000
X=
= 410 = Rs. 14·63.
∑f
1st
10
1200
1200
10
2nd
12
1200
1200
=
12
100
3rd
15
1200
1200
=
15
80
4th
20
1200
1200
=
20
60
5th
24
1200
1200
=
24
50
∑fX = Rs. 6000
Total
= 120
∑f = 410
Remark. For an alternative solution of this problem see Harmonic Mean, Example 5·58.
Example 5·9. For a certain frequency table which has only been partly reproduced here, the mean was
found to be 1·46.
No. of accidents
Frequency (No. of days)
:
:
0
46
1
?
2
?
3
25
4
10
5
5
Total
200
Calculate the missing frequencies.
Solution. Let X denote the number of accidents and let the missing frequencies corresponding to X = 1
and X = 2 be f1 and f2 respectively.
We have
COMPUTATION OF ARITHMETIC MEAN
Frequency
No. of accidents
fX
200 = 86 + f1 + f2
⇒
f1 + f2 = 200 – 86 = 114
–
X =
⇒
⇒
∑fX f1 + 2f2 + 140
=
∑f =
200
…(*)
1·46 (Given)
f1 + 2f2 + 140 = 1·46 × 200 = 292
f1 + 2f2 = 292 – 140 = 152
Subtracting (*) from (**), we get
f2 = 152 – 114 = 38
…(**)
(X)
(f)
—————————————————————————
——————————————————————————
———————————————————————
0
1
2
3
4
5
Total
46
f1
f2
25
10
5
86 + f1 + f2 = 200
0
f1
2f2
75
40
25
f 1 + 2f 2 + 140
5·12
BUSINESS STATISTICS
Substituting in (*), we get
f1 = 114 – f2 = 114 – 38 = 76
Example 5·10. The following are the hourly salaries in rupees of 20 employees of a firm :
130
62
145
118
125
76
151
142
110
65
116
100
103
71
85
80
122
132
98
95
The firm gives bonuses of Rs. 10,15, 20, 25 and 30 for individuals in the respective salary groups
exceeding Rs. 60 but not exceeding Rs. 80, exceeding Rs. 80 but not exceeding Rs. 100, and so on up to
exceeding Rs. 140 but not exceeding Rs. 160. Find the average hourly bonus paid per employee.
Solution. First we shall express the given data in the form of a grouped frequency distribution with
salaries (in Rupees) in the class intervals 61—80, 81—100, 101—120, 121—140 and 141—160. The first
value in the above distribution is 130, so we put a tally mark against the class interval 121—140; next value
is 62, so we put a tally mark against the class 61—80 and so on. Thus the grouped frequency distribution is
as follows :
Salary (in Rs.)
COMPUTATION OF AVERAGE HOURLY BONUS PER EMPLOYEE
Tally Marks
Frequency (f)
Bonus (in Rs.) (X)
fX
61—80
||||
5
10
50
81—100
||||
4
15
60
101—120
||||
4
20
80
121—140
||||
4
25
100
141—160
|||
3
30
90
∑ f = 20
Total
∑ f X = 380
∑ fX 380
=
= Rs. 19.
∑ f 20
Example 5·11. The mean salary paid to 1,000 employees of an establishment was found to be
Rs. 180·40. Later on, after disbursement of salary, it was discovered that the salary of two employees was
wrongly entered as Rs. 297 and 165. Their correct salaries were Rs. 197 and Rs. 185. Find the correct
Arithmetic Mean.
Solution. Let the variable X denote the salary (in rupees) of an employee. Then we are given :
∴
Average hourly bonus paid per employee =
–
∑X
X = 1000
= 180·40
⇒
∑X = 180400
…(*)
Thus the total salary disbursed to all the employees in the establishment is Rs. 1,80,400. After
incorporating the corrections we have :
Corrected ∑X = 180400 – (sum of wrong salaries) + (sum of correct salaries)
= 180400 – (297 + 165) + (197 + 185)
= 180400 – 462 + 382 = 180320
∴
Corrected mean salary =
180320
1000 =
Rs. 180·32.
Example 5·12. The table below shows the number of skilled and unskilled workers in two small
communities, together with their average hourly wages :
Worker category
Skilled
Unskilled
Ram Nagar
Shyam Nagar
Number
Wage per hour
Number
Wage per hour
150
850
Rs. 180
Rs. 130
350
650
Rs. 175
Rs. 125
Determine the average hourly wage for each community. Also give reasons why the results show that
the average hourly wage in Shyam Nagar exceeds the hourly wage in Ram Nagar even though in Shyam
Nagar the average hourly wage of both categories of workers is lower.
5·13
AVERAGES OR MEASURES OF CENTRAL TENDENCY
–
–
Solution. Let n1 and n2 denote the number, andX1 and X2 denote the wages (in rupees) per hour of the
–
skilled and unskilled workers respectively in the community. Let X be the mean wages of all the workers in
the community.
Ram Nagar. We have :
Shyam Nagar. We have :
–
n1 = 150,
X1 = Rs. 180
–
n2 = 850,
–
∴
X1 = 175
–
X2 = Rs. 130
n2 = 650,
X2 = 125
–
–
–
n X + n2X2 350 × 175 + 650 × 125
X = 1 1
=
350 + 650
n1 + n2
–
n X + n 2 X2 150 × 180 + 850 × 130
X = 1 1
=
150 + 850
n1 + n2
–
–
n1 = 350,
∴
+ 110500 137500
= 270001000
=
= Rs. 137·50
1000
=
61250 + 81250 142500
= 1000
1000
= Rs. 142·50
Thus, we see that the average wage per hour for all the workers combined is higher in Shyam Nagar
than in Ram Nagar, although the average hourly wages of both types of workers are lower in Shyam Nagar.
The reasons for this somewhat strange looking result may be assigned as follows :
The difference in the average hourly wages in Ram Nagar and Shyam Nagar :
(a) For skilled workers is Rs. (180 – 175) = Rs. 5
(b) For unskilled workers is Rs. (130 – 125) = Rs. 5
Thus, although the difference in the wages of skilled and unskilled workers in both the communities is
same viz., Rs. 5, the number of skilled workers getting relatively higher wages than the unskilled workers is
much more in Shyam Nagar than in Ram Nagar and the number of unskilled workers getting relatively less
wages is much less in Shyam Nagar than in Ram Nagar. In fact, the ratio of skilled workers to unskilled
workers in Ram Nagar is 150 : 850 i.e., 3 : 17 while in Shyam Nagar, it is 350 : 650 i.e., 7 :13.
Example 5·13. The mean of marks in Statistics of 100 students in a class was 72. The mean of marks of
boys was 75, while their number was 70. Find out the mean marks of girls in the class.
Solution. In the usual notations we are given :
n = 70, –x = 75 ; n + n = 100, –x = 72 ; ∴ n = 100 – 70 = 30. We want –x .
1
1
1
2
–
–
–x = n1 x 1 + n2 x2
We have
n1 + n2
∴
72 × 100 = 5250 + 30 –x2
2
2
70 × 75 + 30 –x 2
72 =
100
–x = 7200 – 5250 = 1950 = 65
⇒
⇒
2
30
30
Hence, the mean of marks of girls in the class is 65.
Example 5·14. The average daily wage of all workers in a factory is Rs. 444. If the average daily
wages paid to male and female workers are Rs. 480 and Rs. 360 respectively, find the percentage of male
and female workers employed by the factory.
Solution. Let n1 and n2 denote respectively the number of male and female workers in the factory and
–
–
–
X1 and X2 denote respectively their average daily salary (in Rupees). Let X denote the average salary of all
the workers in the factory. Then we are given that :
–
–
X1 = 480,
We have
⇒
X2 = 360
–
X = 444
–
n1X1 + n2 X2
X =
n1 + n2
–
–
and
444 (n1 + n2 ) = 480n1 + 360n2
⇒
⇒
–
—
–
(n1 + n2 ) X = n1 X 1 + n2 X2
(480 – 444)n1 = (444 – 360)n2
5·14
BUSINESS STATISTICS
∴
n1 84 7
= =
n2 36 3
⇒
36n1 = 84n2
Hence, the male workers in the factory are :
and the female workers in the factory are :
7
7+3
3
7+3
7
× 100 = 10
× 100 = 70%
3
× 100 = 10
× 100 = 30%.
Example 5·15. The arithmetic mean height of 50 students of a college is 5′—8˝. The height of 30 of
these is given in the frequency distribution below. Find the arithmetic mean height of the remaining 20
students.
Height in inches
:
5′—4˝
5′—6˝
5′—8˝
5′—10˝
6′—0˝
Frequency
:
4
12
4
8
2
Solution. Let the variable X denote the height of the students in inches.
COMPUTATION OF MEAN HEIGHT OF 30 STUDENTS
Height in inches Frequency
fd
d = X –2 68
(f)
(X)
————————————————————————————
——————————————————
——————————————————
—————————————————
64
66
68
70
72
–
X1 = Mean height (in inches) of n1 = 30 students
2 × (–8)
= A + h∑fd
∑f = 68 + 30
= 204030– 16 = 2024
30 inches
…(*)
–2
–1
0
1
2
4
12
4
8
2
∑ f = 30
–
–8
–12
0
8
4
∑ fd = – 8
–
Let X2 denote the mean height of the remaining n2 = 50 – 30 = 20 students. If X is the mean height of
the 50 students, then we are given that :
–
X = 5′—8˝ = 68 ˝
–
–
n1 X1 + n2 X2
X =
n1 + n2
–
∴
⇒
–
2024 + 20 X2 = 68 × 50 = 3400 ⇒
30 ×
68 =
–
X2 =
( 2024
30 ) + 20 X
–
2
[Using (*)]
50
3400 – 2024 1376
= 20 =
20
68·8′′ = 5′—8·8′′.
5·5. WEIGHTED ARITHMETIC MEAN
The formulae discussed so far in (5·1) to (5·6) for computing the arithmetic mean are based on the
assumption that all the items in the distribution are of equal importance. However, in practice, we might
come across situations where the relative importance of all the items of the distribution is not same. If some
items in a distribution are more important than others, then this point must be borne in mind, in order that
average computed is representative of the distribution. In such cases, proper weightage is to be given to
various items - the weights attached to each item being proportional to the importance of the item in the
distribution. For example, if we want to have an idea of the change in cost of living of a certain group of
people, then the simple mean of the prices of the commodities consumed by them will not do, since all the
commodities are not equally important, e.g., wheat, rice, pulses, housing, fuel and lighting are more
important than cigarettes, tea, confectionery, cosmetics, etc.
Let W 1 , W2,…, W n be the weights attached to variable values X 1 , X2 ,…, X n respectively. Then the
–
weighted arithmetic mean, usually denoted by Xw is given by :
–
Xw =
W1 X1 + W2X2 + … + W n Xn ∑WX
=
W1 + W2 + … + W n
∑W
…(5·11)
This is precisely same as formula (5·2) with f replaced by W.
In case of frequency distribution, if f1 , f2 ,…, fn are the frequencies of the variable values X1 , X 2 ,…, Xn
respectively then the weighted arithmetic mean is given by :
5·15
AVERAGES OR MEASURES OF CENTRAL TENDENCY
–
Xw =
W1(f1 X1 ) + W2(f2 X2 ) + … + Wn(fn Xn ) ∑W(f X)
=
W1 + W2 + …+ Wn
∑W
…(5·12)
where W1, W2,…, Wn are the respective weights of X1, X2 ,…, Xn .
Example 5·16. A candidate obtained the following percentages of marks in an examination : English
60; Hindi 75; Mathematics 63; Physics 59 ; Chemistry 55. Find the candidate’s weighted arithmetic mean
if weights 1, 2, 1, 3, 3 respectively are allotted to the subjects.
Solution. Let the variable X denote the percentage of marks in the examination.
COMPUTATION OF WEIGHTED MEAN
Subject
Marks (%) (X)
Weight (W)
English
60
1
60
Hindi
75
2
150
Mathematics
63
1
63
Physics
59
3
177
Chemistry
55
∴
Weighted Arithmetic Mean (in%) =
∑WX 615
∑W = 10
WX
3
165
∑W = 10
∑WX = 615
= 61·5.
Example 5·17. Comment on the performance of the students in three universities given below, using
simple and weighted averages :
University
Course
of study
Bombay
% of pass
Calcutta
No. of students
(in ’00s)
% of pass
Madras
No. of students
(in ’00s)
% of pass
No. of students
(in ’00s)
M.A.
71
3
82
2
81
2
M. Com.
83
4
76
3
76
3·5
B.A.
73
5
73
6
74
4·5
B.Com.
74
2
76
7
58
2
B.Sc.
65
3
65
3
70
7
M.Sc.
66
3
60
7
73
2
Solution.
COMPUTATION OF SIMPLE AND WEIGHTED AVERAGES
University →
Course
of study
M.A.
M.Com.
B.A.
B.Com,
B.Sc.
M.Sc.
Total
Bombay
Pass
% age
Calcutta
Pass
% age
No. of
students
(in ’00s)
Madras
No. of
students
Pass
% age
No. of
students
(in ’00s)
(W3)
(in ’00s)
(X1)
(W1 )
W1X1
(X2)
(W2)
W2X2
(X3)
71
83
73
74
65
66
432
3
4
5
2
3
3
20
213
332
365
148
195
198
1451
82
76
73
76
65
60
432
2
3
6
7
3
7
28
164
228
438
532
195
420
1977
81
76
74
58
70
73
432
2
3·5
4·5
2
7
2
21
W3X3
162
266
333
116
490
146
1513
5·16
BUSINESS STATISTICS
University
Bombay
Calcutta
Madras
Simple Average
Weighted Average
∑X1 432
=
= 72
6
6
∑X2 432
=
= 72
6
6
∑X3 432
=
= 72
6
6
∑W1X1 1451
=
= 72·55
20
∑W1
∑W2 X2 1977
=
= 70·61
28
∑W2
∑W3 X3 1513
=
= 72·05
21
∑W3
On the basis of the simple arithmetic mean which comes out to be same for each University viz., 72, we
cannot distinguish between the pass percentage of the students in the three Universities. However, the
weighted averages show that the results are the best in Bombay University (which has highest weighted
average of 72·55), followed by Madras University (which has the weighted average 72·05), while Calcutta
University shows the lowest performance.
Example 5·18. From the results of two colleges A and B below state which of them is better and why ?
Name of
College A
Examination
College B
Appeared
Passed
Appeared
Passed
300
500
2000
1200
4000
250
450
1500
750
2950
1000
1200
1000
800
4000
800
950
700
500
2950
M.A.
M. Com.
B.A.
B.Com.
Total
Solution.
Name of
Examination
Appeared
(WA)
College A
Passed
M.A.
300
250
M.Com.
500
450
B.A.
2000
1500
B.Com.
1200
750
Total
4000
2950
Pass % age
(XA)
Appeared
(WB )
250
× 100 = 83·33
300
450
× 100 = 90
500
1500
× 100 = 75
2000
750
× 100 = 62·5
1200
2950
× 100 = 73·75
4000
College B
Passed
1000
800
1200
950
1000
700
800
500
4000
2950
Pass % age
(XB)
800
× 100 = 80
1000
950
× 100 = 79·17
1200
700
× 100 = 70
1000
500
× 100 = 62·5
800
2950
× 100 = 73·75
4000
On the basis of the given information, it is not possible to decide which college is better, since the
criterion for ‘better college’ is not defined. Let us try to solve this problem by taking the ‘Higher Pass
Percentage’ as criterion for ‘better college’.
From the calculation table, we find that the pass percentage in M.A., M.Com. and B.A. is better in
college A than in college B and in B.Com. the pass percentage is same in both the colleges. The simple
arithmetic mean of pass percentages in all the four courses is :
–
XA = 83·33 + 90 +4 75 + 62·50 = 310·83
4 = 77·71
;
–
XB = 80 + 79·17 +4 70 + 62·50 = 291·67
4 = 72·92
Since the mean pass percentage is higher for college A than for college B, we are tempted to conclude
that college A is better than college B. However, this conclusion is not valid since the average pass
percentage is affected by the number of students appearing in the examination in different courses. An
appropriate average would be the weighted average of these pass percentages in different courses, the
corresponding weights being the number of students appearing in the examination. The weighted means
are :
–
XW (A) =
∑WAXA
∑WA
Total number of students passed in college A
2950
= Total
number of students appeared in college A × 100 = 4000 × 100 = 73·75
5·17
AVERAGES OR MEASURES OF CENTRAL TENDENCY
–
XW (B) =
∑WB XB
Total number of students passed in college B
∑WB = Total number of students appeared in college B
× 100 = 2950
4000 × 100 = 73·75
On comparing the weighted means, we conclude that both the colleges A and B are equally good on the
basis of the criterion of higher pass percentage for all the students taken together.
Example 5·19. (a). Show that the weighted arithmetic mean of first n natural numbers whose weights
are equal to the corresponding numbers is equal to (2n + 1)/3.
(b) Also obtain simple arithmetic mean.
Solution. The first n natural numbers are 1, 2, 3, …, n.
We know that :
1 + 2 + 3 + … + n = n(n2+ 1)
12 + 2 2 + 3 2 + … + n2 =n(n + 1)6(2n + 1)
}
COMPUTATION OF WEIGHTED A.M.
X
W
WX
——————————————————————
——————————————————————
——————————————————————
…(*)
1 +2 +3 +…+n
X w = ∑WX
∑W = 1 + 2 + 3 + … + n
2
2
3
2
1
2
3
12
22
32
n
n
n2
..
.
(a) Weighted arithmetic mean is given by :
–
1
2
3
..
.
..
.
[From above Table]
= n(n + 1)6(2n + 1) · n(n2+ 1) = 2n3+ 1
[From (*)]
(b) Simple A.M. of first n natural numbers is
–
1 + 2 + 3 +… + n
X = ∑X
n =
n
=
n(n + 1)
n+1
= 2
2n
[From (*)]
EXERCISE 5·1
1. (a) What is a statistical average ? What are the desirable properties for an average to possess ? Mention different
types of averages and state why the arithmetic mean is the most commonly used amongst them.
(b) State two important objects of measures of central value.
[Delhi Univ. B.Com. (Pass), 1997]
Hint. (i) To obtain a single figure which is representative of the distribution.
(ii) To facilitate comparisons.
2. What are ‘measures of location’ ? In what circumstances would you consider them as the most suitable
measures for describing the central tendency of a frequency distribution ?
3. (a) Explain the properties of a good average. In the light of these properties which average do you think is the
best and why ?
(b) What are the criteria of a satisfactory measure of the central tendency ? Discuss the standard measures of
central tendency and say which of these satisfy your criteria.
(c) What do you mean by an ‘Average’ in Statistics. Mention the essentials of a good average.
4. What do you understand by arithmetic mean ? Discuss its merits and demerits. Also state its important
properties.
5. “The figure of 2·2 children per adult female was felt to be in some respects absurd and the Royal Commission
suggested that the middle class be paid money to increase the average to a rounder and more convenient number.”
(Punch)
Commenting on the above statement, discuss the limitations of the arithmetic average. Also point out the
characteristics of a good measure of central tendency.
6. Calculate the average bonus paid per member from the following data :
Bonus (in Rs.)
:
50
60
70
80
90
100
110
No. of persons
:
1
3
5
7
6
2
1
Ans. Rs. 79.60.
7. (a) Peter travelled by car for 4 days. He drove 10 hours each day. He drove : first day at the rate of 45 km. per
hour, second day at the rate of 40 km. per hour, third day at the rate of 38 km. per hour and fourth day at the rate of
37 km. per hour. What was his average speed ?
Ans. 40 km. p.h.
5·18
BUSINESS STATISTICS
(b) Typist A can type a letter in 5 minutes, typist B in 10 minutes and typist C in 15 minutes. What is the average
number of letters typed per hour per typist ?
Ans. Required average = (12 + 6 + 4)/3 = 7·33.
8. (a) A taxi ride in a city costs one rupee for the first kilometre and sixty paise for each additional kilometre. The
cost for each kilometre is incurred at the beginning of the kilometre, so that the rider pays for a whole kilometre. What
is the average cost for 2 43 kilometres ?
4
Ans. Average cost for 2 34 kms = (100 + 60 + 60) × 11
Paise = 80 Paise.
(b) The mean weight of a student in a group of 6 students is 119 lbs. The individual weights of five of them are
115, 109, 129, 117 and 114 lbs. What is the weight of the sixth student ?
Ans. 130 lbs.
9. (a) Average marks in Statistics of 10 students of a class was 68. A new student took admission with 72 marks
whereas two existing students left the college. If the marks of these students were 40 and 39, find the average marks of
the remaining students.
[Delhi Univ.. B.Com. (Pass), 2000]
– (68 × 10) + 72 – 40 – 39
= 74·7 8 marks (approx.).
Hint. x =
10 + 1 – 2
(b) Shri Narendra Kumar has invested his capital in three securities, namely RELIANCE Ltd., TISCO and
SATYAM : Rs. 40,000; Rs. 50,000 and Rs. 80,000 respectively. If he collects dividends of Rs. 10,000 from each
company, compute his average return from three securities.
[Delhi Univ. B.Com. (Pass), 2000]
Total return
3 × 10‚000
Hint. Average rate of return = Total
=
tnvestment 40‚000 + 50‚000 + 80‚000
Ans. 17·65%.
10. (a) Twelve persons gambled on a certain night. Seven of them lost at an average rate of Rs. 10·50 while the
remaining five gained at an average of Rs. 13·00. Is the information given above correct ? If not, why ?
Ans. Information is incorrect.
(b) Goals scored by a hockey team in successive matches are 5, 7, 4, 2, 4, 0, 5, 5 and 3. What is the number of
goals, the team must score in 10th match in order that the average comes to 4 goals per match.
Ans. 5.
(c) The sum of deviations of a certain number of observations measured from 4 is 72 and the sum of the deviations
of the same value from 7 is –3. Find the number of observations and their mean. [Delhi Univ. B.Com. (Hons.), 1997]
Hint. Let n be the number of observations.
–
–
–
If d = X – A, then X = A + ∑d
; ∴ X = 4 + 72
= 7 + (–3)
· . Solving, we get n = 25, X = 6·88.
n
n
n
(d) The daily average sales of a store were Rs. 2,750 for the month of Feb. 1996. During the month, the highest
and the lowest sales were Rs. 8,950 and Rs. 580 respectively. Find the average daily sales if the highest and the lowest
sales are not taken into account.
[Delhi Univ. B.Com. (Hons.), 1997]
Hint and Ans. n = No. of days in month of February of 1996 (Leap Year) = 29
1
Revised mean = Rs. 27
[∑X – 8,950 – 580] = Rs. 271 [ 29 × 2,750 – 8,950 – 580] = Rs. 2,600.74
Ans. Rs. 2,600.74
(e) Two variables x and y are related by : y = (x – 5)/10 and each of them has 5 observations. If the mean of x is 45,
find the mean of y.
[I.C.W.A. (Foundation), Dec. 2006]
–
–
Ans. y = [ x – 5)/10 ] = 4.
11. (a) The following are the daily salaries in rupees of 30 employees of a firm :
91, 139, 126, 119, 100,
87,
65,
77,
99,
95, 108, 127,
86, 148, 116,
76,
69,
88, 112, 118,
89, 116,
97, 105,
95,
80,
86,
106,
93, 135.
The firm gave bonus of Rs. 10, 15, 20, 25, 30, 35, 40, 45 and 50 to employees in the respective salary groups :
exceeding 60 but not exceeding 70, exceeding 70 but not exceeding 80 and so on up to exceeding 140 but not
exceeding 150. Construct a frequency distribution and find out the total daily bonus paid per employee.
Ans. Average daily bonus = Rs. 27·50.
(b) The management of a college decides to give scholarship to the students who have scored marks 70 and above
70 in Business Statistics. The following are the marks scored by II B.Com. students :
5·19
AVERAGES OR MEASURES OF CENTRAL TENDENCY
71
74
73
74
74
76
85
93
86
91
The scholarship payable is given below :
Marks
:
70—75
Scholarship amount (Rs.) :
100
75—80
200
88
94
80—85
300
91
96
94
98
85—90
400
96
88
90—95
500
99
94
95—100
600
Estimate the total scholarship payable and the average scholarship payable.
(Bangalore Univ. B.Com., 1999)
12. A certain number of salesmen were appointed in different territories and the following data were compiled
from their sales reports :
Sales (’000 Rs.) :
4—8
8—12
12—16 16—20 20—24 24—28 28—32 32—36 36—40
No. of salesmen :
11
13
16
14
—
9
17
6
4
If the average sales is believed to be Rs. 19,920, find the missing information.
Ans. Missing Frequency = 10.
13. The mean of the following frequency distribution is 50. But the frequencies f1 and f 2 in classes 20—40 and
60—80 are missing. Find the missing frequencies.
Class
:
0—20
20—40
40—60
60—80
80—100
Frequency
:
17
f1
32
f2
19
Total 120
[Delhi Univ. B.Com. (Pass), 1997]
Ans. f1 = 28, f2 = 24.
14. (a) The average salary of 49 out of 50 employees in a firm is Rs. 100. The salary of the 50th employee is Rs.
97·50 more than the average salary of all the 50 workers. Find the mean salary of all the employees in the firm.
(b) The mean of 99 items is 55. The value of 100th item is 99 more than the mean of 100 items. What is the value
of 100th item.
[Delhi Univ. B.Com (Hons.), 2001]
Ans. (a) Rs. 101·99, (b) 155.
15. (a) The mean of 200 items was 50. Later on it was discovered that two items were wrongly read as 92 and 8
instead of 192 and 88. Find out the correct mean.
Ans. 50·9.
(b) The average daily income for a group of 50 persons working in a factory was calculated to be Rs. 169. It was
later discovered that one figure was mis-read as 134 instead of the correct value 143. Calculate the correct average
income.
Ans. Rs. 169·18.
(c) The average marks of 80 students were found to be 40. Later, it was discovered that a score of 54 was misread
as 84. Find the correct mean of 80 students.
[C.S. (Foundation), June 2001]
Ans. 39·625.
16. 100 students appeared for an examination. The results of those who failed are given below :
Marks
5
10
15
20
25
30
Total
No. of Students
4
6
8
7
3
2
30
If the average marks of all students were 68.6, find out average marks of those who passed.
[Delhi Univ. B.Com. (Hons.), 2008]
—
∑fX 475
Ans. n1 + n2 = 100, n1 = 30 ⇒ n2 = 70 ; X 1 = Mean marks of failed students =
=
·
∑f
30
—
—
—
n 1 X 1 + n2 X 2 475 + 70 X 2
—
=
= 68.6 ⇒ X 2 = 91.21
n 1 + n2
100
Marks
No. of students
17. Fifty students appeared in an examination. The
results of the passed students in given in the adjoining table.
40
6
The average marks of all the students is 52. Find the
50
14
average marks of the students who failed in the examination.
60
7
[I.C.W.A. (Foundation), Dec. 2006]
70
5
80
4
Ans. 21.
90
4
18. Out of 50 examinees, those passing the examination are shown below. If average marks of all the examinees is
5·16, what would be the average marks of examinees having failed in it ?
—
X 12 =
5·20
Marks obtained
No. of students passing the Exam.
BUSINESS STATISTICS
:
:
4
8
5
10
6
9
7
8
9
6
4
3
[C.S. (Foundation), June 2002]
Ans. 2·1.
19. (a) The mean age of a combined group of men and women is 30 years. If the mean age of the group of men is
32 and that of the group of women is 27, find out the percentage of the men and women in the group.
Ans. Men = 60%, Women = 40%.
(b) The mean annual salary of all employees in a company is Rs. 25,000. The mean salary of male and female
employees is Rs. 27,000 and Rs. 17,000 respectively. Find the percentage of males and females employed by the
company.
[C.A. (Foundation), Nov. 1995]
Ans. Males = 80%, Females = 20%.
(c) If the means of two groups of m and n observations are 40 and 50 respectively, and the combined mean of two
groups is 42, find the ratio m : n.
[I.C.W.A. (Foundation), June 2007]
Ans. m : n = 4 : 1.
20. (a) The mean marks obtained by 300 students in the subject of Statistics are 45. The mean of the top 100 of
them was found to be 70 and the mean of the last 100 was known to be 20. What is the mean of the remaining 100
students ?
Ans. 45.
(b) The mean hourly wage of 100 labourers working in a factory, running two shifts of 60 and 40 workers
respectively, is Rs. 38. The mean hourly wage of 60 labourers working in the morning shift is Rs. 40. Find the mean
hourly wage of 40 labourers working in the evening shift.
[Delhi Univ. B.Com. (Pass), 1996]
Ans. Rs. 35.
21. (a) There are there sections in B.Com. 1st year in a certain college. The number of students in each section and
the average marks obtained by them in the Statistics paper in the annual examination are as follows :
Section
Average marks in Statistics
No. of Students
A
75
50
B
60
60
C
55
50
Find the average marks obtained by the students of all the sections taken together.
Ans. 63·125.
(b) B.Com. (Pass) III year has three Sections A, B and C with 50, 40, 60 students respectively. The mean marks
for the three sections were determined as 85, 60 and 65 respectively. However, marks of a student of section A were
wrongly recorded as 50 instead of zero. Determine the mean marks of all the three sections put together.
[Delhi Univ. B.Com. (Pass), 1995]
–
– 50 × 84 + 40 × 60 + 60 × 65
50 × 85 – 50 + 0
Hint. Corrected XA =
= 84 ; ∴ Combined mean (x) =
= 70
50
50 + 40 + 60
22. The mean monthly salary paid to 77 employees in a company was Rs. 78. The mean salary of 32 of them was
Rs. 75 and that of other 25 was Rs. 82. What was the mean salary of the remaining ?
Ans. Rs. 77·80.
23. Define the weighted arithmetic mean of a set of numbers. Show that it is unaffected if all the weights are
multiplied by some common factor.
24. A contractor employs three types of workers-male, female and children. To a male worker he pays Rs. 16 per
hour, to a female worker Rs. 13 per hour and to a child worker Rs. 10 per hour. What is the average wage per hour paid
by the contractor if the number of males, females and children is 20, 15 and 5 respectively ?
Ans. Rs. 14·12.
25. Define a ‘weighted mean’. Under what circumstances would you prefer it to an unweighted mean ?
Calculate the weighted mean price of a table from the following data, assuming that weights are proportional to
the number of tables sold :
Price per table (Rs.)
:
3600
4000
4400
4800
No. of tables sold
:
14
11
9
6
Ans. Rs. 4070.
5·21
AVERAGES OR MEASURES OF CENTRAL TENDENCY
26. Compute the weighted arithmetic mean of the index number from the data below :
Food
125
7
Index No.
Weight
Group
Fuel and Light
141
4
Clothing
133
5
House Rent
173
1
Miscellaneous
182
3
Ans. 141·15.
27. The following table gives the distribution of 100 accidents during seven days of the week of a given month.
During the particular month there are 5 Mondays, Tuesdays and Wednesdays and only four each of the other days.
Calculate the average numbr of accidents per day.
Days
No. of Accidents
Days
No. of Accidents
26
16
12
10
Thursday
Friday
Saturday
8
10
18
Sunday
Monday
Tuesday
Wednesday
Ans. 14·13 –~ 14.
28. To produce a scooter of a certain make, labour of different kinds is required in quantities as follows :
Skilled labour
:
50 hours
Semi-skilled labour
:
100 hours
Unskilled labour
:
300 hours
If hourly wage rates for these three kinds of labour are Rs. 100, Rs. 70 and Rs. 20 respectively, what is the average
labour cost per hour in producing the scooter ?
[Delhi Univ. B.A. (Econ. Hons.), 1990]
Hint. Use weighted arithmetic mean.
Ans. Rs. 40 per hour.
29. A candidate obtained the following percentages of marks in different subjects in the Half-Yearly Examination :
English
Statistics
Cost Accountancy
Economics
Income Tax
46%
67%
72%
58%
53%
It is agreed to give double weights to marks in English and Statistics as compared to other subjects. What is the
simple and weighted arithmetic mean ?
[Delhi Univ. B.Com. (Pass), 2002]
–
—
Ans. X = 59·2% and XW = 58·43%
30. Calculate simple and weighted arithmetic averages from the following data and comment on them :
Designation
Daily salary (in Rs.)
Class I Officers
Class II Officers
Subordinate staff
Clerical staff
Lower staff
Strength of the cadre
1,500
800
500
250
100
10
20
70
100
150
Ans. Simple Arithmetic Mean = Rs. 630. ; Weighted Arithmetic Mean = Rs. 302·86.
31. Comment on the performance of the students of three Universities given below using an appropriate average :
University →
Course of Study
↓
M.A.
M.Com.
M.Sc.
B.Com.
B.Sc.
B.A.
A
% of Pass
81
76
73
58
70
74
B
No. of students
in hundreds
2
3·5
2
2
7
4·5
% of Pass
82
76
60
76
65
73
C
No. of students in
hundreds
2
3
7
7
3
6
% of Pass
No. of students in
hundreds
71
83
66
74
65
73
3
4
3
2
3
5
5·22
BUSINESS STATISTICS
Ans. Simple average (A.M.) of pass percentage is 72% in each case; we are unable to distinguish between the
performance of students in the three universities.
However, on the basis of weighted average of pass percentage, University C (72·55%) is the best followed by
University A (72·05%) and University B (70·61%).
32. From the results of two colleges A and B given below, state which of them is better and why ?
Name of Examination
College A
College B
Appeared
Passed
Appeared
Passed
M.A.
60
50
200
160
M.Com.
100
90
240
190
B.A.
400
300
200
140
B.Com.
240
150
160
100
Total
800
590
800
590
Hint and Ans. Find the weighted average of percentage of passed students (X), the corresponding weights (W)
being the number of students appeared.
–
XW (A) =
———————————
———————————
———————————
———————————
———————————
———————————
———————————
———————————
∑WAXA 590
∑WBXB 590
–
=
× 100 = 73·75 ; XW (B) =
= 800 × 100 = 73·75
∑WA
800
∑WB
Taking ‘higher pass percentage’ as the criterion for better college, both the colleges A and B are equally good.
33. A travelling salesman made five trips in two months. The record of sales is given below :
Trip
The sales manager criticised the salesman’s
performance as not very good since his mean daily
sales were only Rs. 54,000 (2,70,000/5). The
salesman called this an unfair statement for his daily
mean sales were as high as Rs. 55,200 (13,80,000/25).
What does each average mean here ? Which average
seems to be more appropriate in this case ?
No. of days
Value of sales
(in ’00 Rs.)
Sales per day
(in ’00 Rs.)
———————————————
——————————————————————
————————————————————————
———————————————————————
1
2
3
4
5
5
4
3
7
6
3,000
1,600
1,500
3,500
4,200
600
400
500
500
700
25
13,800
2,700
Ans. The Manager obtained the simple arithmetic mean of the sales per day, while the salesman obtained the
weighted arithmetic mean. The latter (weighted average) seems to be more appropriate.
5·6. MEDIAN
In the words of L.R. Connor :
“The median is that value of the variable which divides the group in two equal parts, one part
comprising all the values greater and the other, all the values less than median”. Thus median of a
distribution may be defined as that value of the variable which exceeds and is exceeded by the same
number of observations i.e., it is the value such that the number of observations above it is equal to the
number of observations below it. Thus, we see that as against arithmetic mean which is based on all the
items of the distribution, the median is only positional average i.e., its value depends on the position
occupied by a value in the frequency distribution.
5·6·1. Calculation of Median.
Case (I) : Ungrouped Data. If the number of observations is odd, then the median is the middle value
after the observations have been arranged in ascending or descending order of magnitude. For example, the
median of 5 observations 35, 12, 40, 8, 60 i.e., 8, 12, 35, 40, 60, is 35.
In case of even number of observations median is obtained as the arithmetic mean of the two middle
observations after they are arranged in ascending or descending order of magnitude. Thus, if one more
observation, say, 50 is added to the above five observations then the six observations in ascending order of
magnitude are : 8, 12, 35, 40, 50, 60. Thus,
Median = Arithmetic mean of two middle terms = 12 (35 + 40) = 37·5.
Remark. It should be clearly understood that in case of even number of observations, in fact, any value
lying between the two middle values can serve as a median but it is a convention to estimate median by
taking the arithmetic mean of the two middle values.
5·23
AVERAGES OR MEASURES OF CENTRAL TENDENCY
Case (II) : Frequency Distribution. In case of frequency distribution where the variable takes the
values X1, X2,…, Xn with respective frequencies f1 , f2 ,…, fn with ∑f = N, total frequency, median is the size
of the (N + 1)/2th item or observation. In this case the use of cumulative frequency (c.f.) distribution
facilitates the calculations. The steps involved are :
(i) Prepare the ‘less than’ cumulative frequency (c.f.) distribution.
(ii) Find N/2.
(iii) See the c.f., just greater than N/2.
(iv) The corresponding value of the variable gives median.
The example given below illustrates the method.
Example 5·20. Eight coins were tossed together and the number of heads (X) resulting was noted. The
operation was repeated 256 times and the frequency distribution of the number of heads is given below :
No. of heads (X) :
0
1
2
3
4
5
6
7
8
Frequency (f)
:
1
9
26
59
72
52
29
7
1
Calculate median.
Solution.
Here N = ∑ f = 256,
X
⇒
N
2 =
COMPUTATION OF MEDIAN
f
Less than c.f.
———————————————————
—————————————————
——————————————————————————————————
128
0
1
2
3
4
5
6
7
8
1
9
26
59
72
52
29
7
1
1
1 + 9 = 10
10 + 26 = 36
36 + 59 = 95
95 + 72 = 167
167 + 52 = 219
219 + 29 = 248
248 + 7 = 255
255 + 1 = 256
The cumulative frequency (c.f.) just greater than 128
is 167 and the value of X corresponding to 167 is 4.
Hence, median number of heads is 4.
Case (III) : Continuous Frequency Distribution.
As before, median is the size (value) of the (N + 1)/2th
observation. Steps involved for its computation are :
(i) Prepare ‘less than’ cumulative frequency (c.f.)
distribution.
(ii) Find N/2.
(iii) See c.f. just greater than N/2.
(iv) The corresponding class contains the median value and is called the median class.
The value of median is now obtained by using the interpolation formula :
where
and
(
)
h N
–C
f 2
l is the lower limit of the median class,
f is the frequency of the median class,
h is the magnitude or width of the median class,
N = ∑ f, is the total frequency,
C is the cumulative frequency of the class preceding the median class.
Median = l +
…(5·13)
Remarks 1. The interpolation formula (5·13) is based on the following assumptions :
(i) The distribution of the variable under consideration is continuous with exclusive type classes
without any gaps.
(ii) There is an orderly and even distribution of observations within each class.
However, if the data are given as a grouped frequency distribution where classes are not continuous,
then it must be converted into a continuous frequency distribution before applying the formula. This
adjustment will affect only the value of l in (5·13).
2. Median will be abbreviated by the symbol Md.
5·24
BUSINESS STATISTICS
3. The sum of absolute deviations of a given set of observations is minimum when taken from median.
By absolute deviation we mean the deviation after ignoring the algebraic sign. Thus, if we take the
deviation of the given values of the variable X from an assumed mean A , then X – A may be positive or
negative but its absolute value denoted by | X – A |, read as (X – A) modulus or (X – A) mod is always
positive and we have
∑ f | X – A | > ∑ d | X – Md | or ∑ f | X – Md | < ∑ f | X – A |; A ≠ Md.
i.e., the sum of the absolute deviations about any arbitrary point A is always greater than the sum of the
absolute deviations about the median. For further discussion, see Mean Deviation in Chapter 6 on
Dispersion.
5·6·2. Merits and Demerits of Median.
Merits. (i) It is rigidly defined.
(ii) Median is easy to understand and easy to calculate for a non-mathematical person.
(iii) Since median is a positional average, it is not affected at all by extreme observations and as such
is very useful in the case of skewed distributions (c.f. Chapter 7), J-shaped or inverted J-shaped
distributions (c.f. Chapter 4) such as the distribution of wages, incomes and wealth. So in case of extreme
observations, median is a better average to use than the arithmetic mean since the later gives a distorted
picture of the distribution.
(iv) Median can be computed while dealing with a distribution with open end classes.
(v) Median can sometimes be located by simple inspection and can also be computed graphically. (See
Ogive discussed in § 5·6·4.)
(vi) Median is the only average to be used while dealing with qualitative characteristics which cannot
be measured quantitatively but can still be arranged in ascending or descending order of magnitude e.g., to
find the average intelligence, average beauty, average honesty, etc., among a group of people.
Demerits. (i) In case of even number of observations for an ungrouped data, median cannot be
determined exactly. We merely estimate it as the arithmetic mean of the two middle terms. In fact any value
lying between the two middle observations can serve the purpose of median.
(ii) Median, being a positional average, is not based on each and every item of the distribution. It
depends on all the observations only to the extent whether they are smaller than or greater than it ; the exact
magnitude of the observations being immaterial. Let us consider a simple example. The median value of
35, 12, 8, 40 and 60 i.e., 8, 12, 35, 40, 60
is 35. Now if we replace the values 8 and 12 by any two values which are less than 35 and the values 40
and 60 by any two values greater than 35 the median is unaffected. This property is sometimes described by
saying that median is sensitive.
(iii) Median is not suitable for further mathematical treatment i.e., given the sizes and the median
values of different groups, we cannot compute the median of the combined group.
(iv) Median is relatively less stable than mean, particularly for small samples since it is affected more
by fluctuations of sampling as compared with arithmetic mean.
Example 5·21. (a) In a batch of 15 students, 5 students failed in a test. The marks of 10 students who
passed were 9, 6, 7, 8, 8, 9, 6, 5, 4, 7. What was the median of all the 15 students ?
(b) If the relation between two variables x and y is 2x + 3y = 7, and the median of y is 2, find the
median of x.
[I.C.W.A. (Foundation), Dec. 2005]
Solution. (a) The marks of 10 students who passed when arranged in ascending order of magnitude
are :
4, 5, 6, 6, 7, 7, 8, 8, 9, 9.
Since the five students who failed must have scored less than 4 marks, the marks of 15 students when
arranged in ascending order are :
., ., ., ., ., 4, 5, 6, 6, 7, 7, 8, 8, 9, 9.
(1)
Here N = 15. Hence, the median value is the middle value viz., 8th value in the series (1). Hence,
median is 6.
5·25
AVERAGES OR MEASURES OF CENTRAL TENDENCY
(b) We are given :
1
x = (7 – 3y) …(**)
2
Since the change of origin and the scale in the observations does not result in any change in the order
(rank) of the observations, get from (**) and (*),
1
1
1
Median (x) = [7 – 3 Median (y)] = (7 – 3 × 2) = ·
2
2
2
Median (y) = 2 …(*)
and
2x + 3y = 7
⇒
Example 5·22. The following table shows the age distribution of persons in a particular region.
Age
No. of persons
Age
No. of persons
(years)
(in thousands)
(years)
(in thousands)
Below 10
2
Below 50
14
” 20
5
” 60
15
” 30
9
” 70
15·5
” 40
12
70 and over
15·6
(i) Find the median age.
(ii) Why is the median a more suitable measure of central tendency than the mean in this case ?
Solution.
COMPUTATION OF MEDIAN
Number of persons
c.f.
(i) First of all we shall convert the given distribution into
Age
in ’000 (f)
(less than)
(in years)
the continuous frequency distribution as given in the
adjoining table and then compute the median.
0—10
2
2
—————————————————————
———————————————————————————
——————————————————
N
15·6
Here 2 = 2 = 7·8. Cumulative frequency (c.f.) greater
than 7·8 is 9. Thus the corresponding class 20—30 is the
median class. Hence, using the median formula (5·13), we get
Median = 20 +
10
4
5
(7·8 – 5) = 20 + 2 × 2·8
= 20 + 5 × 1·4 = 27
Hence, median age is 27 years.
10—20
20—30
30—40
40—50
50—60
60—70
70 and over
5–2=3
9–5=4
12 – 9 = 3
14 – 12 = 2
15 – 14 = 1
15·5 – 15 = 0·5
15·6 – 15·5 = 0·1
N = ∑ f = 15·6
5
9
12
14
15
15·5
15·6
(ii) In this case median is a more suitable measure of central tendency than mean because the last class
viz., 70 and over is open end class and as such we cannot obtain the class mark for this class and hence
arithmetic mean cannot be computed.
Example 5·23. The frequency distribution of weight in grams of mangoes of a given variety is given
below. Calculate the arithmetic mean and the median.
Weight in grams
:
Number of mangoes :
410—419
14
420—429
20
430—439
42
440—449
54
450—459
45
460—469
18
470—479
7
Solution. Since the interpolation formula for median is based on continuous frequency distribution we
shall first convert the given inclusive class interval series into exclusive class interval series.
CALCULATIONS FOR MEAN AND MEDIAN
Weight in grams
(Class boundaries)
409·5—419·5
419·5—429·5
429·5—439·5
439·5—449·5
449·5—459·5
459·5—469·5
469·5—479·5
Total
No. of Mangoes (f)
14
20
42
54
45
18
7
∑f = 200 = N
Mid-value (X)
444·5
d = X –10
414·5
424·5
434·5
444·5
454·5
464·5
474·5
–3
–2
–1
0
1
2
3
fd
– 42
– 40
– 42
0
45
36
21
∑fd = –22
(Less than) c.f.
14
34
76
130
175
193
200
5·26
BUSINESS STATISTICS
–
10 × (–22)
h∑fd
Mean (X) = A + N = 444·5 + 200 = 444·5 – 1·1 = 443·4 gms.
N/2 = 100. The c.f. just greater than 100 is 130.
Hence, the corresponding class 439·5— 449·5 is the median class. Using the median formula, we get
Md = l +
( )
h
f
N
2 –C
10
= 439·5 + 54 (100 – 76)
× 24
= 439·5 + 1054
= 439·50 + 4·44 = 443·94 gms.
Example 5·24. Find the missing frequency from the following distribution of daily sales of shops, given
that the median sale of shops is Rs. 2,400.
Sale in hundred Rs. :
0—10
10—20
20—30
30—40
40—50
No. of shops
:
5
25
—
18
7
Solution. Let the missing frequency be ‘a’.
Since median sales is Rs. 2,400 (24 hundred),
20—30 is the median class. Using median formula, we get
24 = 20 +
∴
(
10 55 + a
a
2 – 30
)
⇒4=
10
a
(
55 + a – 60
2
4a = 5a – 25
⇒
a = 25.
Hence, the missing frequency is 25.
)
CALCULATIONS FOR MEDIAN
No. of shops
Cumulative
Sales in
(f)
frequency (c.f.)
hundred Rs.
—————————————————————
————————————————————
——————————————————————
= 5(aa– 5)
0—10
10—20
20—30
30—40
40—50
5
25
a
18
7
5
30
30 + a
48 + a
N = 55 + a
Example 5·25. In the frequency distribution of 100 families given below, the number of families
corresponding to expenditure groups 20—40 and 60—80 are missing from the table. However, the median
is known to be 50. Find the missing frequencies.
Expenditure
No. of families
:
:
0—20
14
20—40
?
40—60
27
60—80
?
80—100
15
Solution. Let the missing frequencies for the classes 20—40 and 60—80 be f1 and f2 respectively.
COMPUTATION OF MEDIAN
No. of families
c.f.
(f)
(Less than)
Expenditure
From the adjoining table, we have
(in Rupees)
∑f = 56 + f1 + f2 = 100
(Given)
0—20
⇒
f1 + f2 = 100 – 56 = 44
…(*)
20—40
Since median is given to be 50, which lies in the 40—60
class 40—60, therefore, 40—60 is the median class. 60—80
Using the median formula, we get :
80—100
—————————————————————
——————————————————————————————
———————————————————
50 = 40 + 20
27 [50 – (14 + f1 )]
∴
20
10 = 27
(36 – f1 )
⇒
27 = 2(36 – f1 ) = 72 – 2f1
⇒
⇒
14
f1
27
f2
15
N = 100 = 56 + f1 + f2
14
14 + f1
41 + f1
41 + f1 + f2
56 + f1 + f2
20
50 – 40 = 27
[36 – f1 ]
2f1 = 72 – 27 = 45
~
⇒
f1 = 45
2 = 22·5 – 23.
[Since frequency can’t be fractional]
Substituting in (*), we get f2 = 44 – f1 = 44 – 23 = 21.
5·6·3. Partition Values. The values which divide the series into a number of equal parts are called the
partition values. Thus median may be regarded as a particular partition value which divides the given data
into two equal parts.
Quartiles. The values which divide the given data into four equal parts are known as quartiles.
Obviously there will be three such points Q1, Q2 and Q3 such that Q1 ≤ Q2 ≤ Q3, termed as the three
quartiles. Q 1 , known as the lower or first quartile is the value which has 25% of the items of the distribution
5·27
AVERAGES OR MEASURES OF CENTRAL TENDENCY
below it and consequently 75% of the items are greater than it. Incidentally Q 2, the second quartile,
coincides with the median and has an equal number of observations above it and below it. Q3, known as the
upper or third quartile, has 75% of the observations below it and consequently 25% of the observations
above it.
The working principle for computing the quartiles is basically the same as that of computing the
median.
To compute Q1, the following steps are required :
(i) Find N/4, where N = ∑f is the total frequency.
(ii) See the (less than) cumulative frequency (c.f.) just greater than N/4.
(iii) The corresponding value of X gives the value of Q 1 . In case of continuous frequency distribution,
the corresponding class contains Q1 and the value of Q1 is obtained by the interpolation formula :
(
)
h N
–C
…(5·14)
f 4
where l is the lower limit , f is the frequency, and h is the magnitude of the class containing Q1,
and
C is the cumulative frequency (c.f.) of the class preceding the class containing Q1.
Similarly to compute Q3, see the (less than) c.f., just greater than 3N/4. The corresponding value of X
gives Q3. In case of continuous frequency distribution, the corresponding class contains Q 3 and the value of
Q3 is given by the formula :
h 3N
Q3 = l + f 4 – C
…(5·15)
Q1 = l +
(
)
where
l is the lower limit, h is the magnitude, and f is the frequency of the class containing Q3,
and
C is the c.f. of the class preceding the class containing Q3.
Deciles. Deciles are the values which divide the series into ten equal parts. Obviously there are nine
deciles D1, D2, D3,…, D9, (say), such that D 1 ≤ D2 ≤ … ≤ D9. Incidentally D5 coincides with the median.
The method of computing the deciles Di, (i = 1, 2,…, 9) is the same as discussed for Q 1 and Q 3 . To
×N
compute the ith decile Di, (i = 1, 2,…, 9) see the c.f. just greater than i 10
·. The corresponding value of X is
Di. In case of continuous frequency distribution the corresponding class contains Di and its value is
obtained by the formula :
h
×N
Di = l + f i 10
– C , (i = 1, 2,…, 9)
…(5·16)
(
)
where
l is the lower limit, f is the frequency and h is the magnitude of the class containing Di,
and
C is the c.f. of the class preceding the class containing Di
Percentiles. Percentiles are the values which divide the series into 100 equal parts. Obviously, there
are 99 percentiles P 1 , P 2 ,…, P99 such that P1 ≤ P 2 ≤ … ≤ P99. The ith percentile Pi, (i = 1, 2,…, 99) is the
×N
value of X corresponding to c.f. just greater than i100
. In case of continuous frequency distribution, the
corresponding class contains P i and its value is obtained by the interpolation formula :
Pi = l +
h
f
(
i×N
100
)
– C , (i = 1, 2,…, 99)
…(5·17)
where
l is the lower limit, f is the frequency and h is the magnitude of the class containing Pi,
and
C is the c.f. of the class preceding the class containing Pi.
In particular, we shall have :
P25 = Q1,
P50 = D5 = Q2,
P75 = Q3,
D1 = P10,
D2 = P20,
D3 = P30,…,
D9 = P90.
Remark. Importance of partition values. Partition values, particularly the percentiles are specially
useful in the scaling and ranking of test scores in psychological and educational statistics. In the data
relating to business and economic statistics, these partition values, specially quartiles, are useful in
personnel work and productivity ratings.
5·28
BUSINESS STATISTICS
5·6·4. Graphic Method of Locating Partition Values. The various partition values viz., quartiles,
deciles and percentiles can be easily located graphically with the help of a curve called the cumulative
frequency curve or Ogive. The procedure involves the following steps :
Less Than Ogive
Steps 1. Represent the given distribution in the form of a less than cumulative frequency distribution.
2. Take the values of the variable (in the case of frequency distribution) and the class intervals (in the
case of continuous frequency distribution) along the horizontal scale (X-axis) and the cumulative frequency
along the vertical scale (Y-axis).
3. Plot the c.f. against the corresponding value of the variable (in the case of frequency distribution)
and against the upper limit of the corresponding class (in the case of continuous frequency distribution).
4. The smooth curve obtained by joining the points so obtained by means of a free-hand drawing is
called ‘less than’ cumulative frequency curve or ‘less than’ ogive.
The various partition values can be easily obtained from this ogive as illustrated in Example 5·32.
More Than Ogive. In this case we form the ‘more than’ cumulative frequency distribution and plot it
against the corresponding value of the variable or against the lower limit of the corresponding class (in case
of continuous frequency distribution). The curve obtained on joining the points so obtained by smooth freehand drawing is called ‘more than’ cumulative frequency curve or ‘more than’ ogive.
Remark. If we draw a perpendicular from the point of intersection of the two ogives on the x-axis, the
foot of the perpendicular gives the value of median.
Example 5·26. The following data gives the distribution of marks of 100 students. Calculate the most
suitable average, giving the reason for your choice. Also obtain the values of quartiles, 6th decile and 70th
percentile from the following data.
Marks
Less than 10
”
20
”
30
”
40
No. of students
5
13
20
32
Marks
Less than 50
”
60
”
70
”
80
No. of students
60
80
90
100
Solution. We are given ‘less than’ cumulative frequency distribution. We shall first convert it into a
grouped frequency distribution. Since
‘marks’ is a discrete random variable COMPUTATIONS FOR MEDIAN, QUARTILES AND PERCENTILES
Frequency (f)
Less than c.f. Class Boundaries
taking only integral values, the
Class
classes are : Less than 10, 10—19,…, Less than 10
5
5
Below 9·5
70—79. Further, since the formulae
10—19
13 – 5 = 8
13
9·5—19·5
for median, quartiles and percentiles
20—29
20 – 13 = 7
20
19·5—29·5
are based on continuous frequency
30—39
32 – 20 = 12
32
29·5—39·5
distribution, we convert the
40—49
60 – 32 = 28
60
39·5—49·5
distribution into exclusive type
50—59
80
–
60
=
20
80
49·5—59·5
classes with class boundaries below
60—69
90
–
80
=
10
90
59·5—69·5
9·5, 9·5—19·5,…, 69·5—79·5 as
70—79
100
–
90
=
10
100
=
N
69·5—79·5
given in the adjoining table.
Since the first class ‘less than 10’ is an open end class, we cannot compute any of the mathematical
averages like mean, geometric mean or harmonic mean. The only averages we can compute in this case are
median and mode. We compute below the median of the above distribution.
Median. 2N = 100
2 = 50. The c.f. just greater than 50 is 60. Hence, the corresponding class 39·5—49·5 is the
median class.
(
)
10 × 18
∴
Median = 39·5 + 10
28 50 – 32 = 39·5 + 28 = 39·50 + 6·43 = 45·93
Hence, median marks are 45·93.
5·29
AVERAGES OR MEASURES OF CENTRAL TENDENCY
3 × 100
3N
100
Quartiles. N
and
= 75. The c.f. just greater than N/4 is 32. Hence, the
4 =
4
4 = 4 = 25
corresponding class 29·5—39·5 contains Q1 which is given by :
10
(
)
Q1 = 29·5 + 12 25 – 20 = 29·5 +
10 × 5
12 =
29·50 + 4·17 = 33·67
The c.f. just greater than 3N/4 = 75 is 80. Hence, the corresponding class 49·5—59·5 contains Q3 which
is given by :
10
(
)
Q3 = 49·5 + 20 75 – 60 = 49·50 +
6N 6 × 100
Decile. 10 = 10 =
10 × 15
20 =
49·5 + 7·5 = 57·0.
6th
60. The c.f. just greater than 60 is 80. Hence, the corresponding class 49·5—59·5
contains D6 which is given by :
10
(
)
D6 = 49·5 + 20 60 – 60 = 49·5
70N
70 × 100
70th Percentile. 100 = 100 = 70. The c.f. just greater than 70 is 80. Hence, the corresponding class
49·5—59·5 contains P70 which is given by :
10
(
)
P70 = 49·5 + 20 70 – 60 = 49·5 +
10 × 10
20 =
49·5 + 5 = 54·5.
Example 5·27. Comment on the following statement :
“The median of a distribution is N/2, the lower quartile is N/4 and the upper quartile in 3N/4”. (Here
N denotes the total frequency.)
Solution. The statement is wrong. The median of a distribution is not N/2 but it is the value of the
variable X which divides the distribution into two equal parts i.e., median is the value of the variable X such
that N/2 (i.e., 50%) of the observations are less than it and N/2 observations exceed it. The lower quartile
Q1 is not N/4 but it is the value of the variable such that N/4 (i.e., 25%) of the observations are less than Q1.
Similarly the upper quartile Q3 is not 3N/4 but it is the value of the variable such that 3N/4 (i.e., 75%) of the
observations are less than it.
Example 5·32. The following are the marks obtained by the students in Statistics :
Marks
Number of students
Marks
Number of students
10 marks or less
20 ”
”
30 ”
”
4
10
30
40 marks or less
50 ”
”
60 ”
”
40
47
50
Draw a ‘less than’ ogive curve on the graph paper and show therein :
(i) The range of marks obtained by middle 80% of the students.
(ii) The median.
Also verify your results by direct formula calculations.
Solution. The above data can be arranged in the form of a continuous frequency distribution as given
in the adjoining table.
Less Than Ogive. Plot the less than c.f. against the
Frequency (f)
(Less than) c.f.
Marks
corresponding value of the variable in the original table
0—10
4
4
(or against the upper limit of the corresponding class in
10—20
10
–
4
=
6
10
the adjoining table) and join these points by a smooth
20—30
30 – 10 = 20
30
free hand curve to obtain ogive. [See Fig. 5·1]
———————————————————————
——————————————————————————
——————————————————————————
N
30—40
40 – 30 = 10
(i) At the frequency 2 = 25, (along the Y-axis)
40—50
47 – 40 = 7
draw a line parallel to x-axis meeting the ogive at point
50—60
50 – 47 = 3
P. Draw PM perpendicular to the x-axis. Then OM = 27·5, is the median marks.
40
47
N = ∑ f = 50
5·30
BUSINESS STATISTICS
(ii) The range of the marks obtained by the
middle 80% of the students is given by P90 – P 10. To
find P 90 and P 10 graphically, at the frequency
90
100
10
N = 45 and 100 N = 5, draw lines parallel to the xaxis meeting the (less than) ogive at Q and R
respectively. Draw QN and RL perpendicular to the xaxis. Then
P90 = ON = 47 (app.) and P10 = OL =11·7 (app.)
∴ Required range of marks
= P90 – P10 = 47 – 11·7 = 35·3.
OGIVE
50
40
Ogive
(less than)
30
P
20
10
Values by Direct Calculations
R
L
M
N
10
20
30
40
50
60
M
=
27·5
P10 = 11·7 (app.) d
P90 = 47 (app.)
0
N
Median. Here 2 = 25. The c.f. just greater than 25
is 30. Thus the corresponding class 20—30 is the
median class. Using median formula, we get
Median = 20 + 10
20
Q
( 502 – 10 ) = 20 + 21 × 15 = 20 + 7·5 = 27·5
10
100
P10 and P90.
Fig. 5·1.
10
N = 100
× 50 = 5
The c.f. greater than 5 is 10. Hence, P10 lies in the corresponding class 10—20.
∴
10
P10 = 10 + 10
6 (5 – 4) = 10 + 6 = 10 + 1·67 = 11·67
90
100 N
90
= 100
× 50 = 45. The c.f. greater than 45 is 47.
Hence, the corresponding class 40—50 contains P90 and
10 × 5
P90 = 40 + 10
7 (45 – 40) = 40 + 7 = 40 + 7·14 = 47·14
Hence, the range of the marks obtained by the middle 80% of the students is
P90 – P10 = 47·14 – 11·67 = 35·47.
Example 5·28. For a group of 5000 workers, the hourly wages vary from Rs. 20 to Rs. 80. The wages
of 4 per cent of the workers are under Rs. 25 and those of 10 per cent are under 30; 15 per cent of the
workers earn Rs. 60 and over, and 5 per cent of them get Rs. 70 and over. The quartile wages are Rs. 40
and Rs. 54, and the sixth decile is Rs. 50. Put this information in the form of a frequency table.
Solution. We are given : N = 5000.
25
(a) Q1 = 40 Rs. ⇒ 25% i.e., 100
× 5000 = 1250 workers earn less than Rs. 40.
(b) D6 = 50 Rs.
⇒ 60% i.e.,
(c) Q3 = 54 Rs.
⇒ 75% i.e.,
60
100 ×
75
100 ×
5000 = 3000 workers earn below Rs. 50.
5000 = 3750 workers earn below Rs. 54.
Further, we are given that :
4
(i) 4% i.e., 100
× 5000 = 200 workers earn under Rs. 25.
10
(ii) 10% i.e., 100 × 5000 = 500 workers earn under Rs. 30.
15
(iii) 15% i.e., 100
× 5000 = 750 workers earn Rs. 60 and over and
5
(iv) 5% i.e., 100
× 5000 = 250 workers earn over Rs. 70.
Using the above information, we can compute the frequencies for the following class intervals :
Wages in Rs. : Under 25, 25—30, 30—40, 40—50, 50—54, 54—60, 60—70, 70 and over,
5·31
AVERAGES OR MEASURES OF CENTRAL TENDENCY
as given in the following table :
Hourly Wages (in Rs.)
Under 25
25—30
30—40
40—50
50—54
54—60
60—70
70 and over
No. of workers
Less than c.f.
200
500
1250
3000
3750
5000 – 750 = 4250
—
—
200
500 – 200 = 300
1250 – 500 = 750
3000 – 1250 = 1750
3750 – 3000 = 750
4250 – 3750 = 500
750 – 250 = 500
250
Since the number of workers with wages under Rs. 25 is
200 and further, since it is given that the wages vary from Rs. 20
to Rs. 80, the first class viz., under 25 can be taken as 20—25
and the last class, viz., 70 and over can be taken as 70—80. In
the above table, the various classes are of unequal widths.
Rearranging and combining them to have classes with equal
magnitude of 10 each, the final frequency distribution of wages
of 5000 workers is as shown in the adjoining Table.
FREQUENCY DISTRIBUTION OF
WAGES OF WORKERS
No. of workers
Hourly wages
(f)
(in Rs.)
——————————————————————————
—————————————————————————————————
20—30
30—40
40—50
50—60
60—70
70—80
200 + 300 = 500
750
1750
750 + 500 = 1250
500
250
EXERCISE 5·2
1. Define median and discuss its relative merits and demerits.
2. The mean is the most common measure of central tendency of the data. It satisfies almost all the requirements of
a good average. The median is also an average, but it does not satisfy all the requirements of a good average. However,
it carries certain merits and hence is useful in particular fields. Critically examine both the averages.
3. What do you understand by central tendency ? Under what conditions is median more suitable than other
measures of central tendency ?
4. In each of the following cases, explain whether the description applies to mean, median or both :
(i) Can be calculated from a frequency distribution with open end classes.
(ii) The values of all items are taken into consideration in the calculation.
(iii) The values of extreme items do not influence the average.
(iv) In a distribution with a single peak and moderate skewness to the right, it is closer to the concentration of
the distribution.
Ans. (i) median,
(ii) mean,
(iii) median
(iv) median.
5. Find the medians of the following two series :
(i)
38
34
39
35
32
31
37
30
41
(ii)
30
31
36
33
29
28
35
36
Ans. (i) 35, (ii) 32.
6. What are the properties of median ?
Following are the marks obtained by a batch of 10 students in a certain class test in Statistics (X) and Accountancy
(Y).
Roll No. :
1
2
3
4
5
6
7
8
X
:
63
64
62
32
30
60
47
46
Y
:
68
66
35
42
26
85
44
80
In which subject is the level of knowledge of the students higher ?
Ans. Md (X) = 46·5, Md. (Y) = 55. Level of knowledge of students is higher in Accountancy.
7. Find mean and median from the data given below :
Marks obtained :
0—10
10—20
No. of students :
12
18
Ans. Mean = 28,
Median = 27·41
20—30
27
30—40
20
40—50
17
9
35
33
10
28
72
50—60
6
5·32
BUSINESS STATISTICS
8. Calculate arithmetic mean and median from the following series :
Income (Rs.)
:
0—5
5—10
10—15
Frequency
:
5
7
10
15—20
8
20—25
25—30
6
4
[C.S. (Foundation), Dec. 2000]
Ans. Arithmetic mean = 14·375 ; Median = 14.
9. For the data given below, find the missing freqeuncy if the Arithmetic Mean is Rs. 33. Also find the median of
the series :
Loss per shop (Rs.) :
0—10
10—20
20—30
30—40
40—50
50—60
No. of shops
:
10
15
30
—
25
20
[C.A. (Foundation), Nov. 2000]
Ans. Missing frequency = 25 ; Median = 33
10. Given below is the distribution of marks obtained by 140 students in an examination.
Marks
: 10—19 20—29 30—39 40—49 50—59 60—69 70—79 80—89 90—99
No. of students :
7
15
18
25
30
20
16
7
2
Find the median of the distribution.
[C.A. PEE-1, May 2004]
Ans. 51.167.
11. Compute median from the following data :
Mid-value
:
115
125
135
145
155
165
175
185
195
Frequency
:
6
25
48
72
116
60
38
22
3
Hint. The class intervals are : 110—120, 120—130,……, 190—200
Ans. Median = 153·79.
12. You are given below a certain statistical distribution :
Value
:
Less than 100
100—200
200—300
300—400
400 and above
Total
Frequency
:
40
89
148
64
39
380
Calculate the most suitable average giving reasons for your choice.
Ans. Md = 241·22.
13. The following table gives the distribution of marks secured by some students in a certain examination :
Marks
:
0—20
21—30
31—40
41—50
51—60
61—70
71—80
No. of Students :
42
38
120
84
48
36
31
Find : (i) Median marks.
(ii) The percentage of failure if minimum for a pass is 35 marks.
Ans. (i) Md = 40·46 (ii) 31·58%.
14. Calculate the median from the following data :
Weight (in gms.) : 410—419 420—429 430—439 440—449
No. of Apples
:
14
20
42
54
450—459
45
460—469
18
470—479
7
[Andhra Pradesh Univ. B.Com., 1999]
Ans. Median = 443·94 gms
15. Given below is the distribution of 140 candidates obtaining marks X or higher in a certain examination (all
marks are given in whole numbers)
Marks (More than) :
10
20
30
40
50
60
70
80
90
100
Freqeuncy
:
140
133
118
100
75
45
25
9
2
0
Calculate the mean and median marks obtained by the candidates.
Ans. Mean = 50·714, Median = 51·167.
16. The following table gives the weekly wages in rupees in a certain commercial organisation.
Weekly wages (’00 Rs.) :
Frequency
:
30—
3
32—
8
34—
24
36—
31
38—
50
40—
61
42—
38
44—
21
46—
12
48—50
2
Find : (i) the median and the first quartile, (ii) the number of wage earners receiving between Rs. 3700 and
Rs. 4700 per week.
Ans. (i) Md = Rs. 4029.51; Q1 = Rs. 3777.42; (ii) 191.
5·33
AVERAGES OR MEASURES OF CENTRAL TENDENCY
17. Define a percentile. Find the 45th and 57th percentiles for the following data on marks obtained by 100
students :
Marks
20—25
25—30
30—35
35—40
40—45
45—50
10
20
20
15
15
20
No. of Students
[C.A. (Foundation), May 1996]
Ans. P45 = 33·75 ; P57 = 37·33.
18. Find :
(a) the 2nd decile, (b) the 4th decile.
(c) the 90th percentile, and
for the data given below, interpreting clearly the significance of each.
Age of Head of Family
(years)
Under 25
25—29
30—34
35—44
45—54
Ans. D2 = 31·94 years,
Number
(in millions)
2·22
4·05
5·08
10·45
9·47
(d) the 68th percentile
Age of Head of
Family (years)
Number
(in millions)
55—64
65—74
75 and over
6·63
4·16
1·66
————————
Total 43·72
————————
D4 = 40·38 years ,
P90 = 67·98 years,
P68 = 52·87 years.
19. Find the (i) Lower quartile, (ii) Upper quartile, (iii) 7th decile, and (iv) 60th percentile,
for the following frequency distribution :
Wages (Rs.)
:
30—40
40—50
50—60
60—70
70—80
80—90
No. of Persons :
1
3
11
21
43
32
Ans. (i) Rs. 67·14,
(ii) Rs. 83·44,
(iii) Rs. 81·56,
90—100
9
(iv) Rs. 78·37.
20. Draw an ogive for the data given below and show how can the value of median be read off from this graph.
Verify your result.
Class Interval
:
0—5
5—10
10—15
15—20
20—25
25—30
Frequency
:
5
10
15
8
7
5
Ans. Median = 13·5 (approx.); By formula, Md = 13·33.
21. Draw a ‘less than ogive’ from the following data and hence find out the value of lower quartile.
Class Interval
:
0—5
5—10
10—20
20—30
30—40
Frequency
:
5
7
15
20
8
40—50
5
Ans. Q1 = 12.
22. The frequency distribution of heights of 100 college students is as follows :
Height (cms.)
:
141—150
151—160
161—170
171—180
Frequency
:
5
16
56
19
181—190
4
Total
100
Draw an ogive (less than or more than type) of this distribution and from the ogive find
(i) the first quartile, (ii) the median,
(iii) the third quartile, and (iv) Inter-quartile Range.
Ans. Q1 = 161·2 cms, Q3 = 170·1 cms,
Median = 165·7 cms,
I.Q. Range = 8.9 cm.
23. The monthly salary distribution of 250 families in a certain locality in Agra is given below :
Monthly
Salary (Rs.)
No. of Families
Monthly
Salary (Rs.)
More than 0
More than 500
More than 1,000
More than 1,500
250
200
120
80
More than 2,000
More than 2,500
More than 3,000
More than 3,500
No. of Families
55
30
15
5
5·34
BUSINESS STATISTICS
Draw a ‘less than’ ogive for the data given above and hence find out :
(i) Limits of the income of middle 50% of the families ; and
(ii) If income-tax is to be levied on families whose income exceeds Rs. 1,800 p.m., calculate the percentage of
families, which will be paying income-tax.
[Delhi Univ. B.Com. (Hons.), 2007]
Ans. (i) Q1 = Rs. 578 (approx.); Q 3 = Rs. 1850
25
(ii) (2000 – 1500) × (2000 – 1800) + 25 + 15 + 10 + 5 = 65
∴
65
Percentage of families paying income tax = 250 × 100 = 26%.
24. Draw a ‘less than’ and ‘more than’ ogive curve for the following data and find median value :
No. of Children
No. of Families
0
150
1
72
2
50
3
28
4
12
5
8
6
5
[Delhi Univ. B.Com. (Pass), 1999]
Hint. Since the number of children is a discrete random variable which can take only positive integer values, the
given frequency distribution can be expressed as grouped frequency distribution with exclusive type classes as given
below.
Variable
Frequency
0—1
150
1—2
72
2—3
50
3—4
28
4—5
12
5—6
8
6—7
5
Ans. Median from ogive = 1·1 (approx.).
25. With the help of given data, find :
(i) Value of middle 50% items;
(ii) Value of exactly 50% item;
(iii) The value of P 40 and D6;
(iv) Graphically with the help of ogive curve, the values of Q 1, Q3, median, P40 and D6 :
Class Interval
Frequencies
10—14
5
15—19
10
Ans. (i) Q3 – Q1 = 29.19 – 19.92 = 9.27;
20—24
15
25—29
20
(ii) Md = Q2 = 25.13 ;
30—34
10
35—39
5
Total
65
[Delhi Univ. B.Com. (Hons.), 2008]
(iii) P40 = 23.17,
D6 = 26.75
26. One hundred and twenty students appeared for a certain test and the following marks distribution was
obtained:
Marks
Students
Find :
:
:
0—20
10
20—40
30
40—60
36
60—80
30
80—100
14
(i) The limits of marks of middle 30% students.
(ii) The percentage of students getting marks more than 75.
(iii) The number of students who fail, if 35 marks are required for passing.
Ans. (i) P35 = 41·1 ; P65 = 61·3 ;
100
(ii) 120
[( ) ]
30
×
20
5 + 14 = 17·9 %;
(iii) 10 + 15
× 30 = 32·5 ~– 33.
20
27. The expenditure of 1,000 families is given as under :
Expenditure (in Rs.) :
40—59
60—79
80—99
No. of families
:
50
?
500
The median for the distribution is Rs. 87. Calculate the missing frequencies.
Ans. 262·5, 137·5 ~– 263, 137.
28. An incomplete frequency distribution is given as follows :
Variable :
10—20
20—30
30—40
40—50
50–60
Frequency :
12
30
?
65
?
You are given that median value is 46.
(a) Using the median formula, fill up the missing frequencies.
(b) Calculate the Arithmetic Mean of the completed table.
Ans. (a) 34, 45
(b) 45·96
100—119
?
60—70
25
70—80
19
120—139
50
Total
230
5·35
AVERAGES OR MEASURES OF CENTRAL TENDENCY
29. An incomplete distribution is given below :
Variable
:
0—10
10—20
Frequency :
10
20
20—30
?
30—40
40
40—50
?
50—60
25
60—70
15
(i) You are given that the median value is 35. Find out missing frequency (given the total frequency = 170).
(ii) Calculate the arithmetic mean of the completed table.
[Himachal Pradesh Univ. B.Com., 1999; Kerala Univ. B.Com., 1999]
Ans. (i) 35,25
(ii) 35·88.
30. The data in the adjoining table represent travel
expenses (other than transportation) for 7 trips made
during November by a salesman for a small firm :
An auditor criticised these expenses as excessive,
asserting that the average expense per day is Rs. 10
(Rs. 70 divided by 7). The salesman replied that the
average is only Rs. 4·20 (Rs. 105 divided by 25) and that
in any event the median is the appropriate measure and is
only Rs. 3. The auditor rejoined that the arithmetic mean
is the appropriate measure, but that the median is Rs. 6.
You are required to :
Trip
Days
Expenses
(Rs.)
Expenses per day
(Rs.)
—————————————————
————————————————
——————————————————
———————————————————————————
1
2
3
4
5
6
7
0·5
2·0
3·5
1·0
9·0
0·5
8·5
13·50
12·00
17·50
9·00
27·00
9·00
17·00
27
6
5
9
3
18
2
—————————————————
———————————————
——————————————————
——————————————————————————
Total
25·0
105·00
70
(i) Explain the proper interpretation of each of the four averages mentioned.
(ii) Which average seems appropriate to you ?
31. For a certain class of workers, numbering 700, hourly wages vary between Rs. 30 and Rs. 75. 12% of the
workers are earning less than Rs. 35 while 13% are getting equal to or more than Rs. 60, out of which 6% are earning
between Rs. 70 and Rs. 75. The first quartile and median wages are, respectively, Rs. 40 and Rs. 47. The 40th and 65th
percentiles are Rs. 43 and Rs. 53 respectively. You are required to put the above information in the form of a frequency
distribution and estimate the mean wages of the workers.
Ans. Hourly wages (Rs.) :
30—35
No. of workers
:
84
–
X = Rs. 48·33.
35—
91
40—
105
43—
70
47—
105
53—
91
60—
49
32. For a certain group of saree weavers of Varanasi, the median and quartile earnings per hour are Rs. 44·3,
Rs. 43·0 and Rs. 45·9 respectively. The earnings for the group range between Rs. 40 and Rs. 50. Ten per cent of the
group earn under Rs. 42; 13% earn Rs. 47 and over, and 6% Rs. 48 and over. Put these data in the form of a frequency
distribution and obtain the value of the mean wage.
Ans. Hourly Wages (Rs.) :
40—42
No. of workers
:
10
Mean = Rs. 44·50.
42—
15
43—
25
44·3—
25
45·9—
12
47—
7
48—50
6
5·7. MODE
Mode is the value which occurs most frequently in a set of observations and around which the other
items of the set cluster densely. In other words, mode is the value of a series which is predominant in it. In
the words of Croxton and Cowden, “The mode of a distribution is value at the point around which the items
tend to be most heavily concentrated. It may be regarded as the most typical of a series of values.”
According to A.M. Tuttle, ‘Mode is the value which has the greatest frequency density in its immediate
neighbourhood’. Accordingly mode may also be termed as the fashionable value (a derivation of the
French word ‘la Mode’) of the distribution.
In the following statements :
(i) average size of the shoe sold in a shop is 7,
(ii) average height of an Indian (male) is 5 feet 6 inches (1·68 metres approx.),
(iii) average collar size of the shirt sold in a ready-made garment shop is 35 cms,
(iv) average student in a professional college spends Rs. 2,500 per month;
5·36
BUSINESS STATISTICS
Frequency
the average referred to is neither mean nor median but
mode, the most frequent value in the distribution. For
example, by the first statement we mean that there is
maximum demand for the shoe of size No. 7.
5·7·1. Computation of Mode. In case of a frequency
distribution, mode is the value of the variable
corresponding to the maximum frequency. This method
can be applied with ease and simplicity if the distribution
is ‘unimodal’, i.e., if it has only one mode. In other
O
words, this method can be used with convenience if there
is only one value with highest concentration of
observations. For example, in the distribution :
X:
1
2
3
4
5
f:
3
1
18
25
40
Mode
X
Fig. 5·2.
6
30
7
22
8
10
9
6
the maximum frequency is 40 and therefore, the corresponding value of X viz., 5 gives the value of mode.
In case of a frequency curve (see Fig. 5.2) mode corresponds to the peak of the curve.
In the case of continuous frequency distribution, the class corresponding to the maximum frequency is
called the modal class and the value of mode is obtained by the interpolation formula :
Mode = l +
where
and
h (f1 – f0 )
h (f1 – f0 )
=l+
(f1 – f0 ) – (f2 – f1 )
2 f1 – f0 – f2
…(5·18)
l is the lower limit of the modal class,
f1 is the frequency of the modal class,
f0 is the frequency of the class preceding the modal class,
f2 in the frequency of the class succeeding the modal class,
h is the magnitude of the modal class.
The symbols f0 , f1 and f2 can be explained easily as follows :
f0 : Frequency of preceding class,
f1 : Maximum frequency (Frequency of Modal class),
f2 : Frequency of succeeding class.
Remarks 1. It may be pointed out that the formula (5·18) for computing mode is based on the
following assumptions :
(i) The frequency distribution must be continuous with exclusive type classes without any gaps. If the
data are not given in the form of continuous classes, it must first be converted into continuous classes
before applying formula (5·18).
(ii) The class intervals must be uniform throughout i.e., the width of all the class intervals must be the
same. In case of the distribution with unequal class intervals, they should be made equal under the
assumption that the frequencies are uniformly distributed over all the classes, otherwise the value of mode
computed from (5·18) will give misleading results.
2. However, the above technique of locating mode is not practicable in the following situations :
(i) If the maximum frequency is repeated or approximately equal concentration is found in two or more
neighbouring values.
(ii) If the maximum frequency occurs either in the very beginning or at the end of the distribution.
(iii) If there are irregularities in the distribution i.e., the frequencies of the variable increase or decrease
in a haphazard way.
In the above situations mode (or modal class in the case of continuous frequency distribution) is
located by the method of grouping as discussed in Examples 5·31 and 5·32.
5·37
AVERAGES OR MEASURES OF CENTRAL TENDENCY
3. If the method of grouping gives the modal class which does not correspond to the maximum
frequency f1 i.e., the frequency of modal class is not the maximum frequency, then in some situations we
may get 2f1 – f0 – f2 = 0. [This will not be possible if f1 is maximum and f0 and f2 are less than f1 ]. In such a
situation viz., 2f1 – f0 – f2 = 0, the value of mode cannot be computed by the formula.
h (f1 – f0 )
Mode = l +
2f1 – f0 – f2
as it gives
Mode = l + ∞ = ∞.
[·. · 2f1 – f0 – f2 = 0]
In such cases, the value of mode can be obtained by the formula :
h⎟ f1 – f0 ⎟
Mode = l +
…(5·18a)
⎟ f1 – f0 ⎟ + ⎟ f1 – f2 ⎟
where | A | represents the absolute (positive) value of A.
Formula (5·18a) is only an approximate formula and does not give very correct result because further
grouping of classes, say, 4 at a time may give different value of the modal class and as such a different
result.
As an illustration, for the following data :
X :
f :
10—20
4
20—30
6
30—40
5
40—50
10
50—60
20
60—70
22
70—80
24
80—90
6
90—100
2
100—110
1
the usual method of grouping (up to 3 classes at a time) will give 60—70 as the modal class such that :
f1 = 22, f0 = 20, f2 = 24 and therefore, 2f1 – f0 – f2 = 44 – 20 – 24 = 0. Hence, usual formula for mode cannot
be applied. Using (5·18a), an approximate value of mode may be obtained as :
10 | 22 – 20 |
×2
Mo = 60 +
= 60 + 10
2 + 2 = 60 + 5 = 65
| 22 – 20 | + | 22 – 24 |
5·7·2. Merits and Demerits of Mode.
Merits. (i) Mode is easy to calculate and understand. In some cases it can be located merely by
inspection. It can also be estimated graphically from a histogram (c.f. § 5·7·3).
(ii) Mode is not at all affected by extreme observations and as such is preferred to arithmetic mean
while dealing with extreme observations.
(iii) It can be conveniently obtained in the case of open end classes which do not pose any problems
here.
Demerits. (i) Mode is not rigidly defined. It is ill-defined if the maximum frequency is repeated or if
the maximum frequency occurs either, in the very beginning or at the end of the distribution; or if the
distribution is irregular. In these cases, its value is located by the method of grouping (c.f. Examples, 5·31).
If the grouping method also gives two values of mode, then the distribution is called bi-modal distribution
(c.f. Example 5·32). We may also come across distributions with more than two modes, in which case it is
called multimodal distribution. In case of bimodal or multimodal distributions, mode is not a representative
measure of location and its estimate is obtained by the empirical relation :
Mode = 3 Median – 2 Mean, discussed in § 5·8.
(ii) Since mode is the value of X corresponding to the maximum frequency, it is not based on all the
observations of the series. Even in the case of the continuous frequency distribution [c.f. Formula (5·18)],
mode depends on the frequencies of modal class and the classes preceding and succeeding it.
(iii) Mode is not suitable for further mathematical treatment. For example, from the modal values and
the sizes of two or more series, we cannot find the mode of the combined series.
(iv) As compared with mean, mode is affected to a greater extent by the fluctuations of sampling.
Uses. Being the point of maximum density, mode is specially useful in finding the most popular size in
studies relating to marketing, trade, business and industry. It is the appropriate average to be used to find
the ideal size e.g., in business forecasting, in the manufacture of shoes or readymade garments, in sales, in
production, etc.
5·38
BUSINESS STATISTICS
5·7·3. Graphic Location of Mode. Mode can be located graphically from the histogram of frequency
distribution by making use of the rectangles erected on the modal, pre-modal and postmodal classes. The
method consists of the following steps :
(i) Join the top right corner of the rectangle erected on the modal class with the top right corner of the
rectangle erected on the preceding class by means of a straight line.
(ii) Join the top left corner of the rectangle erected on the modal class with the top left corner of the
rectangle erected on the succeeding class by a straight line.
(iii) From the point of intersection of the lines in steps (i) and (ii) above, draw a perpendicular to the
X-axis (the horizontal scale). The abscissa (X-coordinate) of the point where this perpendicular meets the
X-axis gives the modal value.
5·8. EMPIRICAL RELATION BETWEEN MEAN (M), MEDIAN (Md) AND MODE (Mo)
In case of a symmetrical distribution mean, median and mode coincide i.e., Mean = Median = Mode
(c.f. Chapter 7 on Skewness). However, for a moderately asymmetrical (non-symmetrical or skewed)
distribution, mean and mode usually lie on the two ends and median lies in between them and they obey the
following important empirical relationship, given by Prof. Karl Pearson.
Mode = Mean – 3 (Mean – Median)
…(5·19)
⇒
Mean – Mode = 3 (Mean – Median)
⇒
Mean – Median = 31 (Mean – Mode)
…(5·19a)
Thus we see that the difference between mean and mode is three times the difference between mean
and median. In other words, median is closer to mean than mode. The above relation between mean (M),
median (Md) and mode (Mo) can be exhibited diagrammatically as follows (Fig. 5.3) :
Remarks. 1. Equation (5·19) may be rewritten to give :
Mode = Mean – 3 Mean + 3 Median
⇒
Mode = 3 Median – 2 Mean
…(5·20)
This formula is specially useful to
RELATIONSHIP BETWEEN ARITHMETIC MEAN,
determine the value of mode in case it is illMEDIAN AND MODE
defined, e.g., in the case of biomodal or
multimodal distributions [c.f. Example 5·39].
Divides area
in halves
2. If we know any two of the three
values M, Md and M o , the third can be Under peak
Centre of
of curve
estimated by using (5·20). The value so
gravity
computed will be more or less same as
obtained by using the exact formula provided
the distribution is moderately asymmetrical.
3. For a positively skewed distribution
Mo MdM
[c.f. Chapter 7], mean will be greater than
median and median will be greater than
Fig. 5·3.
mode i.e.,
M > Md > Mo
⇒
Mo < Md < M
However, in a negatively skewed distribution the order of the magnitudes of the three averages will be
reversed i.e., for negatively skewed distribution, we have
Mo > Md > M
⇒
M < Md < Mo
Example 5.29. (a) Find the mode of the following distribution :
7, 4, 3, 5, 6, 3, 3, 2, 4, 3, 4, 3, 3, 4, 4, 2, 3
[I.C.W.A. (Foundation), June 2005]
(b) If the relation between two variables x and y be 2x + 5y = 24 and mode of y be 4, find the mode
of x.
[I.C.W.A. (Foundation), June 2006]
5·39
AVERAGES OR MEASURES OF CENTRAL TENDENCY
Solution. (a) The frequency distribution of the variable (x) is obtained as given below.
x
2
3
4
5
6
7
Tally Marks
||
|||| ||
||||
|
|
|
Frequency (f)
2
7
5
1
1
1
Since the maximum frequency (7) corresponds to x = 3, the value of mode is 3.
(b) We are given :
1
Mode (y) = 4 …(*)
and
2x + 5y = 24
⇒
x = (24 – 5y) … (**)
2
If x and y are corrected by the relation y = ax + b, then the frequencies of the variable y are the same as
the frequencies of the corresponding values of the variable x. Hence, the modes of the variables x and y are
also connected by the same equation, i.e.,
y = ax + b ⇒ Mode (y) = Mode (ax + b) = a [Mode (x)] + b
…(***)
Hence on using (***), we get from (**),
Mode (x) = Mode
[
]
1
(24 – 5y) = 12 – 52 · Mode (y) = 12 – 52 × 4 = 2
2
[From (*)]
Example 5·30. Find the value of mean , mode and median from the data given below :
Weight (in kg.) :
No. of students :
93—97
3
98—102
5
103—107
12
108—112
17
113—117
14
118—122
6
123—127
3
128—132
1
Solution. Since the formula for mode requires the distribution to be continuous with ‘exclusive type’
classes we first convert the classes into class bounderies as given in the following table :
Weight
(in kg)
93—97
98—102
103—107
108—112
113—117
118—122
123—127
128—132
COMPUTATION OF MEAN, MODE AND MEDIAN
Class
Mid-value
No. of
X – 110
d= 5
boundaries
(X)
students (f)
–3
3
95
92·5—97·5
–2
5
100
97·5—102·5
–1
12
105
102·5—107·5
0
17
110
107·5—112·5
1
14
115
112·5—117·5
2
6
120
117·5—122·5
3
3
125
122·5—127·5
4
1
130
127·5—132·5
N = ∑f = 61
fd
–9
–10
–12
0
14
12
9
4
∑fd = 8
Less than
c.f.
3
8
20
37
51
57
60
61
5×8
Mean. Mean = A + h∑fd
N = 110 + 61 = 110·66 kgs.
Mode. Here maximum frequency is 17. The corresponding class 107·5—112·5 is the modal class.
Using the mode formula, we get
h(f1 – f0 )
– 12)
= 107·5 + 2 5× ×17(17
– 12 – 14
1 – f0 – f2
107·5 + 25
8 = 107·5 + 3·125 = 110·625
Mode = l + 2f
=
kgs.
Median. (N/2) = (61/2) = 30·5. The c.f. just greater than 30·5 is 37. Hence, the corresponding class
107·5—112·5 is the median class. Using the median formula, we get
5
Md = 107·5 + 17
(30·5 – 20) = 107·5 + 5 ×1710·5
= 107·50 + 3·09 = 110·59 kg.
Example 5·31. Construct a frequency distribution showing the frequencies with which words of
different number of letters occur in the extract reproduced below (omitting punctuation marks, treating as
the variable the number of letters in each word) and obtain the median and the mode of the distribution.
5·40
BUSINESS STATISTICS
A candidate at the time of applying for registration as a student of the institute should be not less than
eighteen years of age and have passed the intermediate examination of a university constituted by law in
India or an examination recognized by Central Government as equivalent thereto, or the National Diploma
in Commerce Examination or the Diploma in Rural Service Examination conducted by the National
Council of Rural Higher Education.
Solution. Here the variable X represents the number of letters in each word. For example, in the word
‘candidate’ there are 9 letters c, a, n, d, i, d, a, t and e. Hence X, corresponding to the word ‘candidate’ is 9.
Thus replacing each word by the number of letters in it, the distribution of the number of letters in each
word in the given paragraph is as follows :
1,
4,
2,
8,
9,
4,
5,
11,
2,
8,
2,
2,
3,
5,
2,
3,
4,
2,
11,
7,
2,
3,
10,
2,
8,
3,
2,
5,
3,
4,
3,
7,
12,
6,
7,
11,
2,
3,
10,
8,
1,
12,
2,
2,
7,
11,
10,
3,
2,
2,
7,
8,
3,
1,
2,
7,
9,
10,
3,
2,
6,
11,
8,
5,
COMPUTATION OF MEDIAN
Frequency
(f)
The above data can be arranged in the No. of letters in a Tally Marks
word (X)
form of a frequency distribution as given in the
adjoining table .
1
|||
N 72
2
|||| |||| |||| ||||
Median. Here 2 = 2 = 36. Since c.f. just
3
|||| |||| ||
greater than 36 is 38, the corresponding value
2,
2,
7,
6,
3,
3,
2,
9
(Less than)
c.f.
3
22
34
38
42
45
52
58
61
65
70
72
————————————————————————
—————————————————————————
——————————————————
——————————————————
4
5
6
7
8
9
10
11
12
of X is median, which is 4.
Mode : Since the above frequency
distribution is not regular, the value of mode is
located by the method of grouping.
As the distribution is not regular we cannot
say that the value of mode is 2, which
corresponds to the maximum frequency is 19.
Here we try to locate mode by the method of
grouping as explained in table below.
X
(1)
1
3
2
19
3
12
4
4
5
4
6
3
7
7
8
6
9
3
10
4
11
5
12
2
||||
||||
|||
|||| ||
|||| |
|||
||||
||||
||
COMPUTATION OF MODE : GROUPING TABLE
Frequencies
(2)
(3)
(4)
}
}
}
}
}
}
22
16
7
13
7
7
}
}
}
}
}
}
}
}
}
34
31
8
10
11
16
9
9
3
19
12
4
4
3
7
6
3
4
5
2
11
(5)
}
}
}
35
14
13
(6)
}
}
}
20
16
12
The frequencies in column (1) are the original frequencies. Column (2) is obtained by combining the
frequencies two by two in column (1). Column (3) is obtained on combining the frequencies two by two in
column (1) after leaving the first frequency. If we leave the first two frequencies and combine the
5·41
AVERAGES OR MEASURES OF CENTRAL TENDENCY
frequencies two by two in column (1), we shall get a repetition of values obtained in column (2). Hence, we
proceed to combine the frequencies in column (1) three by three to get column (4). The combination of
frequencies three by three after leaving the first frequency and first two frequencies in column (1) results in
columns (5) and (6) respectively. If we combine the frequencies three by three after leaving the first three
frequencies in column (1), we get a repetition of values obtained in column (4). The maximum frequency in
each column is represented by ‘‘bold type’.
For computing the value of mode we prepare the following analysis table :
ANALYSIS TABLE
Column
No. in above Table
(I)
(1)
(2)
(3)
(4)
(5)
(6)
Frequency of the variable (X)
Maximum
frequency
(II)
19
22
31
34
35
20
Value or combination of values of X corresponding to maximum
frequency in column (II)
(III)
2
2
1
2
3
2
3
1
2
3
4
3
4
5
2
5
4
2
1
Since the value 2 is repeated maximum number (5) of times, the mode is 2.
Example 5·32. Calculate mode from the following data :
Marks
Below 10
20
”
30
”
40
”
50
”
No. of Students
4
6
24
46
67
Marks
Below 60
70
”
80
”
90
”
No. of Students
86
96
99
100
Solution. Since we are given the cumulative frequency
distribution of marks, first we shall convert it into the frequency
distribution as given in the adjoining table.
Further, since the frequencies first decrease, then increase and
again decrease, the distribution is irregular and hence the modal class
is located by the method of grouping as explained in the table given
below.
Frequency (f)
4
6–4 = 2
24 – 6 = 18
46 – 24 = 22
67 – 46 = 21
86 – 67 = 19
96 – 86 = 10
99 – 96 = 3
100 – 99 = 1
Marks
0—10
10—20
20—30
30—40
40—50
50—60
60—70
70—80
80—90
———————————————————————
——————————————————————————
Marks
(1)
0—10
4
10—20
2
20—30
30—40
40—50
50—60
60—70
70—80
80—90
18
22
21
19
10
3
1
(2)
}
}
}
}
6
40
40
13
GROUPING TABLE
Frequencies
(3)
(4)
}
}
}
}
}
}
}
24
20
43
29
62
14
4
(5)
}
}
42
50
(6)
}
}
61
32
5·42
BUSINESS STATISTICS
For computing modal class, we prepare the analysis table as given below :
Column
Maximum
frequency
No. in above Table
(I)
(II)
22
(1)
40
(2)
43
(3)
62
(4)
50
(5)
61
(6)
Number of times the class occurs
ANALYSIS TABLE
Class (es) corresponding to maximum frequency in (II)
(III)
30—40
30—40
30—40
30—40
20—30
30—40
5
20—30
2
40—50
40—50
40—50
40—50
40—50
5
50—60
50—60
50—60
60—70
3
1
In the above table there are two classes viz., 30—40 and 40—50 which are repeated maximum number
(5) of times and as such we cannot decide about the modal class. Thus, even the method of grouping fails to
give the modal class.
We say that in the above example mode is ill-defined and we locate it by the empirical formula :
Mo = 3Md – 2M
…(*)
For computation of Mean (M) and Median (Md), see calculation Table on page 5.52
Mean = A +
h∑fd
10 × (–28)
=
N = 45 + 100
45 – 2·8 = 42·2.
Here 2N = 100
2 = 50. Since c.f. just greater than 50 is 67, the corresponding class 40—50 is the median
class. Hence, using the median formula, we get
10
Median = 40 + 21
Marks
0—10
10—20
20—30
30—40
40—50
50—60
60—70
70—80
80—90
(
100
2
)
– 46 = 40 + 1021× 4 = 40 + 1·9 = 41·9.
COMPUTATION OF ARITHMETIC MEAN AND MEDIAN
Mid-value
Frequency
Less than
X – 45
d = 10
(X)
(f)
c.f.
–4
4
4
5
–3
6
2
15
–2
24
18
25
–1
46
22
35
0
67
21
45
1
86
19
55
2
96
10
65
3
99
3
75
4
100
1
85
∑f = 100
fd
–16
– 6
–36
–22
0
19
20
9
4
∑fd = –28
Substituting the values of M and Md in (*), we get
Mode = 3 × 41·9 – 2 × 42·2 = 125·7 – 84·4 = 41·3.
Example 5.33. In 500 small scale units, the return on investment ranged from 0 to 30 per cent, no unit
sustaining any loss. Five per cent of the units had returns ranging from zero per cent to 5 per cent, 15 per
cent of the units earned returns between 5 per cent and 10 per cent. The median rate of return was 15 per
cent and the upper quartile was 20 per cent. The uppermost layer of returns of 25—30 per cent was earned
by 50 units. Put this information in the form of a frequency table and find the rate of return around which
there is maximum concentration of units.
[Delhi Unit. B.Com. (Hons.) (External), 2007]
Solution. On the basis of the given information we have : N = Total number of units = 500
5
(1) Number of units with returns ranging from 0 to 5% = 5% of 500 =
× 500 = 25
100
5·43
AVERAGES OR MEASURES OF CENTRAL TENDENCY
(2) Number of units with returns between 5% and 10% = 15% of 500 =
15
× 500 = 75
100
500
= 250 units have return ≤ 15%
…(i)
2
∴ Number of units with return exceeding 10% but not exceeding 15% = 250 – (25 + 75) = 150
3N 3 × 500
Upper Quartile (Q3) = 20% ⇒
=
= 375 units have returns ≤ 20%
4
4
∴ Number of units with returns exceeding 15% but ≤ 20% = 375 – 250 = 125
[using (i)]
Number of units with returns between 25% to 30% = 50 (Given)
Hence, by the residual balance, the number of units with returns between 20% to 25%
= 500 – [25 + 75 + 150 + 125 + 50] = 500 – 425 = 75
(3) Median rate of return = 15%
⇒
50% of N =
Thus, the given information can be summarised in
the form of a frequency distribution as given in the
adjoining table.
The rate of return about which there is maximum
concentration of units is given by the Mode of the rate
of returns.
Since maximum frequency is 150, the modal class
is 10—15.
Return in %
No. of units (f)
0—5
5—10
25
75 (f0)
10—15
150 (f1)
15—20
20—25
25—30
125 (f2)
75
50
Modal
Class
h (f1 – f0 )
5 (150 – 75)
5 × 75
= 10 +
= 10 +
= 10 + 3.75 = 13.75
2f1 – f0 – f2
300 – 75 – 125
100
Hence, the rate of return around which there is maximum concentration of units is 13.75%.
Example 5·34. Below is given the frequency distribution of weights of a group of 60 students of a class
in a school :
∴ Mode = l +
Weight in kg.
30—34
35—39
40—44
45—49
Number of students
3
5
12
18
Weight in kg.
50—54
55—59
60—64
Number of students
14
6
2
(a) Draw histogram for this distribution and find the modal value.
(b) (i) Prepare the cumulative frequency (both less than and more than types) distribution, and
(ii) represent them graphically on the same graph paper. Hence, find the (iii) median, and
(iv) co-efficient of quartile deviation.
(c) With the modal and the median values as obtained in (a) and (b), use an appropriate empirical
formula to find the arithmetic mean of this distribution.
(d) If students with weight below 40 kg. are eliminated from the frequency distribution, what will be the
revised mean ? [Calculate the mean of the two rejected classes only and use the result obtained in (c).]
Less than c.f. More than c.f.
Number of
Solution. (a) To draw the histogram and Weight in
students (f)
kgs.
cumulative frequency curves (both less than
and more than types) we first convert the 29·5—34·5
3
3
60
5
distribution into continuous class intervals as 34·5—39·5
8
57
12
given in the adjoining table.
39·5—44·5
20
52
—————————————————————
———————————————————————
—————————————————————————
——————————————————————
44·5—49·5
49·5—54·5
54·5—59·5
59·5—64·5
18
14
6
2
38
52
58
60
40
22
8
2
5·44
BUSINESS STATISTICS
(a) Histogram. Histogram is obtained on erecting rectangles on the class intervals with heights
proportional to the corresponding class frequencies. [See Fig. 5·4.] Mode = OA = 48.
(b) (i) The ‘less than’ and ‘more than’ cumulative frequency distributions are given in the Table in Part
(a).
(ii) The ‘less than’ and ‘more than’ (ogives) are drawn in the Fig. 5·5.
OGIVES
Y
Frequency
60
50
HISTOGRAM
Y
18
16
14
12
10
8
6
4
2
Less than
Ogive
Q 3 = 52·0 (app.)
S
40
30
20
Q2 = 47·8
Q1 = 42·9
10
A
O
29·5 34·5 39·5 44·5 49·5 54·5 59·5 64·5
O0
X
34·5
Class Intervals
3N
4
P
More than
Ogive
R
N
4
N
2
Q M
L
39·5 44·5 49·5 54·5 59·5 64·5
Class Intervals
Fig. 5·4.
X
Fig. 5·5.
(iii) From the point of intersection P of the two curves (ogives), draw perpendicular on X-axis meeting
X-axis at Q. Then OQ = 47·8 kgs. gives the median weight.
(iv) Draw lines parallel to X-axis at frequency equal to N/4 and 3N/4 meeting the less than ogive at
points R & S respectively. From R & S draw perpendiculars to X-axis meeting OX at L, M respectively.
Then Q1 = OL = 42·9 kgs. and Q3 = OM = 51·75 kgs. The Coefficient of Quartile Deviation is given by :
Q –Q
Coefficient of Q.D. = Q 3 + Q1
3
1
51·75 – 42·90
8·85
= 51·75 + 42·90 = 94·65
= 0·0935
(c) The empirical relation between mean, median and mode is given by : Mo = 3Md – 2M
∴
Mean (M) =
3 × 47·8 – 48
3Md – Mo
=
= 143·42 – 48 = 95·4
2 = 47·700 kgs.
2
2
–
(d) Let X1 be the mean of the two classes
Weight in kgs. Mid-value (X)
with weight below 40 kgs.
(f)
fX
———————————————————————
———————————————————————
——————————————————————————
———————————————————
∴
–
X1 =
∑ fX 281
=
= 35·125 kgs.
∑f 8
30—34
35—39
32
37
3
5
96
185
∑ f = 8 = n1, (say)
∑ fX = 281
–
Let X2 be the mean of the distribution obtained on eliminating the first two classes (i.e., classes with
weight below 40 kgs.). Then in the usual notations, we have
–
n1 = 8, X1 = 35·125; n2 = 60 – 8 = 52,
–
Using the formula,
–
X =
–
X2 = ?,
–
X = 47·700
[From Part (c)]
–
n1 X1 + n2 X2
n1 + n 2
, we get
–
60 × 47·700 = 8 × 35·125 + 52 X2
⇒
–
X2 = 286252– 281 = 2581
52 = 49·635 kgs.
5·45
AVERAGES OR MEASURES OF CENTRAL TENDENCY
EXERCISE 5·3
1. What do you understand by mode ? Discuss its relative merits and demerits as a measure of central tendency.
Also give two practical situations where you will recommend the use of mode.
2. What are the desideratta of a good average ? Compare the mean, the median and the mode in the light of these
desideratta. Why are averages called measures of central tendency ?
3. How would you account for the predominant choice of arithmetic mean of statistical data as a measure of
central tendency ? Under what circumstances would it be appropriate to use mode or median ?
4. Point out the merits and demerits of the mean the median, and the mode as measure of central tendency of
numerical data.
5. Compare, giving illustrations, the arithmetic mean, the median and the mode in regard to :
(a) the effect of extreme items in computation,
(b) ease in computation,
(c) stability in sampling situations,
(d) existence of the average as an actual case, and
(e) popular use.
6. (a) Define mode. When is mode said to be ill-defined ?
[Delhi Univ. B.Com. (Pass), 1997]
(b) A stockist of readymade garments should follow which type of average and why ?
[Delhi Univ. B.Com. (Pass), 2001]
7. The Bharat Ball Bearings Ltd., has collected the following data.
12, 19, 21, 30, 13, 19, 22, 31, 17, 20, 24, 31, 18, 21, 27, 31.
(i) Compute the arithmetic mean, the median and the mode using the sixteen observations given.
(ii) Why is the mode said to be an erratic measure of central tendency ?
(iii) Why is the median called a position average ?
Ans. A.M. = 22·25, Md = 21, Mo = 31
8. Calculate mean, median and mode from the following data of the heights in inches of a group of students :
61, 62, 63, 61, 63, 64, 64, 60, 65, 63, 64, 65, 66, 64
Now suppose that a group of students whose heights are 60, 66, 59, 68, 67, and 70 inches, is added to the original
group. Find mean, median and mode of the combined group.
Ans. First group :
M = 63·2, Md = 63·5, Mo = 64
Combined group :
M = 63·75, Md = 64, Mo = 64.
9. Atul gets a pocket money allowance of Rs. 12 per day. Thinking that this was rather less, he asked his friends
about their allowances and obtained the following data which includes his allowance also — (amounts in Rs.)
12, 18, 10, 5, 25, 20, 20, 22, 15, 10, 10, 15, 13, 20, 18, 10, 15, 10, 18, 15, 12, 15, 10, 15, 10, 12, 18, 20, 5, 8.
He presented these data to his father and asked for an increase in his allowance as he was getting less than average
amount. His father, a statistician, countered pointing out that Atul’s allowance was actually more than the average
amount.
Reconcile these statements.
Ans. Atul computed A.M. and his father computed Mode.
10. The number of fully formed apples on 100 plants were counted with following results :
No. of apples
0
1
2
3
4
5
6
7
8
9
10
No. of plants
2
5
7
11
18
24
12
8
6
4
3
(i) How many apples were there in all ?
(ii) What was the average of number of apples per plant ?
(iii) What was the modal number of apples ?
[Delhi Univ., B.Com., 1989, Allahabad Univ. B.Com., 1996]
–
Ans. (i) 486 (ii) X = 4·86, (iii) Mo = 5.
11. Given below is the frequency distribution of marks obtained by 90 students. Compute the arithmetic mean,
median and mode.
5·46
BUSINESS STATISTICS
Marks
No. of students
6
15—19
14
20—24
12
25—29
10
30—34
10
35—39
9
40—44
Ans. Mean = 37·17, Md = 36, Mo = 23·5.
12. Find out the median and mode from the following table :
No. of days absent
Less than 5
Less than 10
Less than 15
Less than 20
Less than 25
No. of students
29
224
465
582
634
Marls
45—49
50—54
55—59
60—64
65—69
No. of students
9
10
5
4
1
No. of days absent
Less than 30
Less than 35
Less than 40
Less than 45
No. of students
644
650
653
655
Ans. Md =12·75, Mo = 11·35.
13. Find out the Mean, Median and the Mode in the following series—
Size (below) :
5
10
15
20
Frequency
:
1
3
13
17
25
30
35
27
36
38
(Andhra Pradesh Univ. B.Com., 1998)
Ans. Mean = 19·74, Md = 21, Mo = 24·3.
14. In 500 small scale industrial units, the return on investment ranged from 0 to 30%, no unit sustaining any loss.
5% of industrial units had returns exceeding 0% but not exceeding 5%. 15% of units had returns exceeding 5% but not
exceeding 10%. Median and upper quartile rate of return was 15% and 20% respectively. The uppermost layer of
returns exceeding 25% but not exceeding 30% was earned by 25%. Present this information in the form of frequency
table with intervals as follows :
Exceeding 0% but not exceeding 5%
;
Exceeding 5% but not exceeding 10%
Exceeding 10% but not exceeding 15% ;
Exceeding 15% but not exceeding 20%
Exceeding 20% but not exceeding 25% ;
Exceeding 25% but not exceeding 30%.
Use N/4, 2N/4, 3N/4 as ranks of lower, middle and upper quartiles respectively. Find the rate of return around
which there is maximum concentration of units.
[Delhi Univ. B.Com. (Hons.), 2008]
Return in %
No. of units
0—5
25
5–10
75
10—15
150
15—20
125
20–25
0
25—30
125
Mode = 13.75; Rate of return around which there is maximum concentration of units is 13.75%.
15. Calculate the arithmetic mean and the median of the frequency distribution given below. Hence calculate the
mode using the empirical relation between the three.
Class limits : 130—134
135—139
140—144
145—149
150—154
155—159 160—164
Frequency :
5
15
28
24
17
10
1
————————
Ans. M = 145·35,
Md = 144·92,
Mo = 144·06.
16. (a) Briefly explain the role of grouping and analysis table in calculation of mode.
[Delhi Univ. B.Com. (Pass), 1999]
(b) From the following data of weight of 122 persons determine the modal weight by the method of grouping.
Weight (in lbs.)
100—110
110—120
120—130
130—140
140—150
No. of persons
4
6
20
32
33
150—160
160—170
170—180
17
8
2
[Osmania Univ. B.Com. 1998]
Hint. Method of grouping gives two modal classes 130—140 and 140—150 i.e., the distribution is bimodal.
Locate the value of mode by using the empirical relation Mo = 3Md – 2M.
Ans. Mean (M) = 139·51 ;
Median (Md) = 139·69;
Mode (Mo) = 140·05.
5·47
AVERAGES OR MEASURES OF CENTRAL TENDENCY
17. Calculate the Mode, Median and Arithmetic average from the following data.
Class
0—2
2—4
4—10
10—15
15—20
20—25
f
8
12
20
10
16
25
Class
25—30
30—40
40—50
50—60
60—80
80—100
f
45
60
20
13
15
4
Hint. Rewrite the frequency distribution with classes of equal magnitude 10.
Ans. Mo = 28·15, Md = 28·29, Mean = 30·08.
18. In the following data, two class frequencies are missing.
Class
100—110
110—120
120—130
130—140
140—150
Frequency
4
7
15
?
40
Class
150—160
160—170
170—180
180—190
190—200
Frequency
?
16
10
6
3
However, it was possible to ascertain that the total number of frequencies was 150 and that the median has been
correctly found to be 146·25.
You are required to find out with the help of the information given :
(i) Two missing frequencies.
(ii) Having found the missing frequencies, calculate arithmetic mean.
(iii) Without using the direct formula, find the value of the mode.
Ans. (i) 24, 25 ;
–
(ii) X = 147·33;
(iii) Mode = 144·08
19. The median and mode of the following hourly wage distribution are known to be Rs. 33·5 and Rs. 34
respectively. Three frequency values from the table are, however, missing. You are required to find out those values.
60—70
Wages in Rs.
:
0—10
10—20
20—30
30—40
40–50
50—60
Total
4
No. of persons :
4
16
?
?
?
6
230
Ans. 60, 100, 40.
20. You are given the following incomplete frequency distribution. It is known that the total frequency is 1000 and
that the median is 413·11. Estimate by calculation the missing frequencies and find the value of the mode.
Value (X)
300—325
325—350
350—375
375—400
Frequency (f)
5
17
80
?
Value (X)
400—425
425—450
450—475
475—500
Frequency (f)
326
?
88
9
Ans. Missing frequencies are 227 and 248 respectively. Mo = 413·98.
21. “Hari put the jar of water and the packet of sweets on the ground and sat down in the shade of the tree and
waited.”
Prepare a frequency distribution for the words in the above sentence taking the number of letters in words as the
variable. Calculate the mean, median and mode.
Ans. Mean = 3·56,
Median = Mode = 3.
22. Treating the number of letters in each word in the following passage as the variable x, prepare the frequency
distribution table and obtain its mean, median, mode.
“The reliability of data must always be examined before any attempt is made to base conclusions upon them. This
is true of all data, but particularly so of numerical data, which do not carry their quality written large on them. It is a
waste of time to apply the refined theoretical methods of Statistics to data which are suspect from the beginning.”
Ans. Mean = 4·565,
Median = 4,
Mode = 3.
5·48
BUSINESS STATISTICS
23. The frequency distribution of marks obtained by 60 students of a class in a college is given below :
Marks
:
30—34
35—39
40—44
45—49
50—54
55—59
60—64
No. of Students :
3
5
12
18
14
6
2
(i) Draw Histogram for this distribution and find the modal value.
(ii) Draw a cumulative frequency curve and find the marks limits of the middle 50% students.
[Delhi Univ. B.Com. (Hons.), 1991]
Ans. (i) Mode = 47·5 marks,
(ii) Q1 = 42·5 marks,
Q3 = 52 marks.
24. Determine the values of Median and Mode of the following distribution graphically. Verify the results by
actual calculations. After verifying, calculate the value of Mean and sketch a curve indicating the general shape of the
distribution and comment.
Size
Frequency
10—19
11
20—29
19
70—79
80—89
90—99
6
3
1
[Delhi Univ. B.Com. (Hons.), 2009]
Hint. Change classes into class boundaries for Md and Mode. Use Ogive for Md and Histogram for Mode
graphically.
Using Formula; Md = 37.83, Mo = 32.35; Mean = [ (3Md – Mo)/2] = 40.57
M > Md > Mo
30—39
21
40—49
16
50—59
10
60—69
8
⇒ Distribution is positively skewed.
25. In a moderately skewed distribution :
(a) Arithmetic mean = 24·6 and the mode = 26·1. Find the value of the median and explain the reason for the
method employed.
(b) In a moderately asymmetrical distribution the value of median is 42·8 and the value of mode is 40. Find the
mean.
(c) In a moderately asymmetrical distribution the value of mean is 75 and value of mode is 60. Find the value of
median.
[Delhi Univ. B.Com. (Pass), 1996]
Ans. (a) Median = 25·1, (b) Mean = 44·2, (c) Median = 70.
26. Find out the missing figures :
(a) Mean = ? (3 Median – Mode)
;
(c) Median = Mode + ? (Mean – Mode) ;
Ans. (a) 1/2, (b) 3, (c) 2/3, (d) 3.
(b) Mean – Mode = ? (Mean – Median)
(d) Mode = Mean – ? (Mean – Median).
27. (a) Which average would you use in the following situations :
(i) Sale of shirts : 16′′,
15 12′′,
15′′,
15′′,
14′′,
13′′,
15′′.
(ii) Marks obtained : 10, 8, 12, 4, 7, 11 and X, (X < 5). Justify your answer.
Ans. (i) Mode,
(ii) Median
(b) A.M. and Median of 50 items are 100 and 95 respectively. At the time of calculations two items 180 and 90
were wrongly taken as 100 and 10. What are the correct values of Mean and Median ?
Ans. Mean = 103·2;
Median is same viz., 95.
(c) Can the values of mean, mode and median be same ? If yes, state the situation.
Ans. M = Md = Mo, for symmetrical distribution.
28. (a) Find out the missing figure ; Mean = ? (3 Median – Mode)
Ans. 1/2.
(b) In a moderately asymmetrical distribution, the values of mode and median are 20 and 24 respectively. Locate
the value of mean.
Ans. 25.
29. Fill in the blanks :
(i) …… can be calculated from a frequency distribution with open end classes.
(ii) In the calculation of …… , all the observations are taken into consideration.
(iii) …… is not affected by extreme observations.
(iv) Average rainfall of a city from Monday to Saturday is 0·3 inch. Due to heavy rainfall on Sunday, the
average rainfall for the week increases to 0·5 inch. The rainfall on Sunday was … …
AVERAGES OR MEASURES OF CENTRAL TENDENCY
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
(xiii)
(xiv)
(xv)
(xvi)
(xvii)
(xviii)
(xix)
(xx)
5·49
The sum of squared deviations is minimum when taken from ……
The sum of absolute deviations is minimum when taken from ……
Median = …… Quartile.
Mean is …… by extreme observations.
Median is the average suited for …… classes.
For studying phenomenon like intelligence and honesty …… is a better average to be used while for
phenomenon like size of shoes or readymade garments the average to be preferred is …… .
Typist A can type a sheet in 5 minutes, typist B in 6 minutes and typist C in 8 minutes. The average number
of sheets typed per hour per typist is ……
The mean of 10 observations is 20 and median is 15. If 5 is added to each observation, the new mean is
…… and median is ……
A distribution with two modes is called …… and with more than two modes is called …… .
Average suited for a qualitative phenomenon is …… .
If 25% of the observations lie above 80, 40% of the observations are less than 50 and 70% are greater than
40, then.
…… = 80 ; …… = 50 ; …… = 40
Relationship between Md, Q1, Q2 and Q3 is ……
D5, P80, Md, D7 and P50 are related by ……
Relationship between D4, Q2, P60, P75 and Q3 is ……
The empirical relationship between mean, median and mode for a moderately asymmetrical distribution is
……
If the maximum frequency is repeated then mode is located by the method of ……
Ans.
(i) Md or Mo (ii) Mean
(iii) Md or Mo
(iv) 1·7′′
(v) Mean
(vi) Median
(vii) Second
(viii) Very much affected
(ix) Open end
(x) Median, Mode
(xi) 9·47
(xii) 25, 20
(xiii) Bi-modal, Multi-modal (xiv) Median
(xv) Q 3 = 80, P 40 = D4 = 50, P30 = 40 (xvi) Q1 ≤ Q2 = Md ≤ Q3 (xvii) D5 = P50 = Md ≤ D7 ≤ P80
(xviii) D4 ≤ Q2 ≤ P60 ≤ P75 = Q3
(xix) Mo = 3Md – 2M.
(xx) Grouping.
30. State, giving reasons, the average to be used in the following situations :
(i) To determine the average size of the shoe sold in a shop.
(ii) To determine the size of agricultural holdings.
(iii) To determine the average wages in an industrial concern.
(iv) To find the per capita income in different cities.
(v) To find the average beauty among a group of students in a class.
Ans. (i) Mode;
(ii), (iii) and (v) Median ;
(iv) Mean.
5·9. GEOMETRIC MEAN
The geometric mean, usually abbreviated as G.M.) of a set of n observations is the nth root of their
product. Thus if X1 , X2 ,…, Xn are the given n observations then their G.M. is given by
n
G.M. = √
X1 × X2 × X3 × … × Xn = (X1 . X2 … Xn )1/n
…(5·21)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
If n = 2 i.e., if we are dealing with two observations only then G.M. can be computed by taking the
square root of their product. For example, G.M. of 4 and 16 is ⎯√⎯⎯⎯
4 × 16 = ⎯⎯
√64 = 8.
But if n, the number of observations is greater than 2, then the computation of the n th root is very
tedious. In such a case the calculations are facilitated by making use of the logarithms. Taking logarithm of
both sides in (5·21), we get
log (G.M.) = 1n log (X1 X2 …Xn )
= 1n (log X1 + log X2 + … + log Xn )
= 1n ∑ log X
…(5·21a)
Thus we see that the logarithm of the G.M. of a set observations is the arithmetic mean of their
logarithms.
5·50
BUSINESS STATISTICS
Taking Antilog of both sides in (5·21a), we finally obtain,
G.M. = Antilog
[ 1n ∑ log X ]
…(5 ·21b)
In case of frequency distribution (Xi, fi); i = 1, 2, …, n, where the total number of observations is
N = ∑f,
[
]
1/N
G.M. = (X1 × X1 × …f1 times) × (X2 × X2 × …f2 times) × … × (Xn × Xn × … fn times
1/N
= (X 1 f1 × X2 f2 × … × Xn fn)
…(5·22)
Taking logarithm of both sides in (5·22), we get
[ (
1
= N [ log X f
1
= N [ f log X
1
log G.M. = N log X1 f1 . X2 f2 … Xn fn
1 1
1
=
⇒
1
]
]
log X ]
+ log X2 f2 + … + log Xn fn
+ f2 log X2 + … + fn
n
1
∑ f log X
N
G.M. = Antilog
[ 1N ∑ f log X ]
… (5·22a)
… (5·22b)
In the case of grouped or continuous frequency distributions, the values of X are the mid-values of the
corresponding classes.
Steps for the Computation of G.M. in (5·22b)
1. Find log X, where X is the value of the variable or the mid-value of the class (in case of grouped or
continuous frequency distribution).
2. Compute f × log X i.e., multiply the values of log X obtained in step 1 by the corresponding
frequencies.
3. Obtain the sum of the products f log X obtained in step 2 to get ∑ f log X.
4. Divide the sum obtained in step 3 by N, the total frequency.
5. Take the Antilog of the value obtained in step 4. The resulting figure gives the value of G.M.
5·9·1. Merits and Demerits of Geometric Mean.
Merits : (i) Geometric mean is rigidly defined.
(ii) It is based on all the observations.
(iii) It is suitable for further mathematical treatment. If G1 and G 2 are the geometric means of two
groups of sizes n1 and n2 respectively, then the geometric mean G of the combined group of size
n1 + n2 is given by
n log G1 + n2 log G2
log G = 1
…(5·23)
n1 + n2
Remark. The result in (5·23) can be easily generalised to the case of k groups as follows :
If G 1 , G2, …,G k are the geometric means of the k groups of sizes n1 , n2 ,…, nk respectively, then the
geometric mean G of the combined group of size n1 + n2 + … + nk is given by :
log G
=
n1 log G1 + n2 log G2 + … + nk log Gk
n1 + n2 + … + n k
…(5·23a)
(iv) Unlike arithmetic mean which has a bias for higher values, geometric mean has bias for smaller
observations and as such is quite useful in phenomenon (such as prices) which has a lower limit (prices
cannot go below zero) but has no such upper limit.
(v) As compared with mean, G.M. is affected to a lesser extent by extreme observations.
(vi) It is not affected much by fluctuations of sampling.
AVERAGES OR MEASURES OF CENTRAL TENDENCY
5·51
Demerits. (i) Because of its abstract mathematical character, geometric mean is not easy to understand
and to calculate for a non-mathematical person.
(ii) If any one of the observations is zero, geometric mean becomes zero and if any one of the
observations is negative, geometric mean becomes imaginary regardless of the magnitude of the other
items.
Uses. In spite of its merits and limitations, geometric mean is specially useful in averaging ratios,
percentages, and rates of increase between two periods. For example, G.M. is the appropriate average to be
used for computing the average rate of growth of population or average increase in the rate of profits, sales,
production, etc., or the rate of money.
Geometric mean is used in the construction of Index Numbers. Irving Fisher’s ideal index number is
based on geometric mean [See Chapter 10 on Index Numbers].
While dealing with data pertaining to economic and social sciences, we usually come across the
situations where it is desired to give more weightage to smaller items and small weightage to larger items.
G.M. is the most appropriate average to be used in such cases.
5·9·2. Compound Interest Formula. Let us suppose that P0 is the initial value of the variable (i.e., the
value of the variable in the beginning and Pn be its value at the end of the period n and let r be the rate of
growth per unit per period.
Since r is the rate of growth per unit per period, growth for period 1 is P0 r and thus the value of the
variate at the end of period 1 is r P 0 + P0 = P0 (1 + r). For the 2nd period the initial value of the variable
becomes P0 (1 + r). The growth for the 2nd period is P0 (1 + r) r and consequently the value of the variable
at the end of 2nd period is
P2 = P0 (1 + r) + P0 (1 + r)r = P0 (1 + r) [1 + r] = P0 (1 + r) 2
Similarly proceeding, the value of the variable at the end of period 3 is
P3 = P0 (1 + r)2 + P0 (1 + r)2 r = P0 (1 + r)2 [1 + r] = P0 (1 + r)3
and finally, its value at the end of period n will be given
Pn = P0 (1 + r)n ,
…(5·24)
which is the compound interest formula for money.
Equation (5·24) involves four unknown quantities :
Pn : The value at the end of period n ;
n : The length of the period ;
P0 : The value in the beginning ;
r : The rate per unit per period.
If we are given P0 , r and n we can compute Pn by using (5·24) directly. However, (5·24) can be used to
obtain any one of the four values when the remaining three values are given. For example, for given P n , r
and n we have :
P
P0 = (1 +nr)n
…(5·24a)
5·9·3. Average Rate of a Variable Which Increases by Different Rates at Different Periods. Let us
suppose that instead of the values of the variable increasing at a constant rate in each period, the rate per
unit per period is different, say, r1, r 2 ,…,rn for the 1st, 2nd,…and nth period respectively. Then, as
discussed in the previous section we shall get :
P1 = The value at the end of 1st period = P0 (1 + r1 )
P2 = The value at the end of 2nd period = P0 (1 + r1 ) (1 + r2 )
Pn = The value at the end of period n = P0(1 + r1 ) (1 + r2 )…(1 + rn )
…(*)
If r is assumed to be the constant rate of growth per unit per period, then we get, [From (*)],
Pn = P0 (1 + r) n
…(**)
5·52
BUSINESS STATISTICS
Hence, equating the values of Pn in (*) and (**), the average rate of growth over the period n is given
by :
(l + r)n = (l + r1 ) (l + r2)…(l + rn )
[
⇒
l + r = (l + r1 ) (l + r2)…(l + rn )
]
1/n
…(5·25)
If r1 , r2 ,…, rn denote the percentage growth per unit per period for the n periods respectively then we
have
1/n
r1
r2
r
r
1 + 100
= 1+
1+
…
1+ n
…(5·26)
100
100
100
[(
)(
)]
) (
where r is the average percentage growth rate over n periods.
∴
[
]
= [ (100 + r ) (100 + r ) … (100 + r ) ]
100 + r = (100 + r1 ) (100 + r2 ) … (100 + rn )
⇒
r
1
2
1/n
1/n
n
– 100
…(5·26a)
Thus we see that if rates are given as percentages, then the average percentage growth rate can be
obtained on subtracting 100 from the G.M. of (100 + r1 ), (100 + r2 ),…,(100 + rn ).
Remark. It should be clearly understood that average percentage growth rate is given by (5·26) and not
by the geometric mean of r1 , r2 ,…, rn.
5·9·4. Wrong Observations and Geometric Mean. Let us suppose that the value of the geometric
mean computed from n observations, say, X1 , X 2 ,…,Xn is G. On checking, it is found that some of the
observations, say, X1, X2 and X3 were wrongly copied instead of the correct observations X 1 ′, X2 ′ and X3 ′.
We are interested in computing the correct value of the geometric mean.
G = Geometric Mean of X1 , X2 , …,Xn = (X1 . X2 . X3 …Xn )1/n.
…(***)
On replacing the wrong observations X 1 , X 2 and X3 by the correct values, the corrected value of the
geometric mean, say, G′ is given by :
1/n
X′ X′ X′
G′ = (X 1 ′ X2 ′ X3 ′ … X n )1/n = X1 · X2 · X3 · X1 X2 X3 … Xn
(
(
1
2
)
3
)
(
)
X1 ′ X2 ′ X3 ′ 1/n
X1 ′ X2 ′ X3 ′ 1 /n
=
G.
·
[From (***)]
(5·26b)
X1 X1 X3
X1 X2 X3
The result in (5·26b), can be generalised to the case of more than three observations. For illustration,
see Example 5·37.
= (X1 X2 X3 … Xn )1/n
Example 5·35. (a) Find the Geometric Mean of 2, 4, 8, 12, 16 and 24.
(b) If the observations 2, 4, 8 and 16 occur with frequencies 4, 3, 2 and 1 respectively, find their
geometric mean.
[I.C.W.A. (Foundation), Dec. 2005]
Solution. (a)
X
log X
2
0·3010
log (G.M.)
∴
Aliter
∴
4
= n1 ∑
0·6021
log X
= 5·4697
6 =
8
12
16
24
0·9031
1·0792
1·2041
1·3802
0·9116
Total
5·4697
[Using (5·21 a)]
G.M. = Antilog (0·9116) = 8·158
G.M. = (2 × 4 × 8 × 12 × 16 × 24)1/6 = (294912)1/6
log G.M. = 16 log 294912 = 5·4698
= 0·9116
6
(b) We are given :
⇒
G.M. = Antilog (0·9116) = 8·158.
x
2
4
8
16
f
4
3
2
1
N = ∑f = 10
5·53
AVERAGES OR MEASURES OF CENTRAL TENDENCY
G.M. = (24 × 43 × 82 × 161 )1/10 = [24 × (22 )3 × (23 )2 × 24]1/10
= [2 4 × 2 6 × 2 6 × 2 4 ] 1/10 = [2 4 + 6 + 6 + 4 ] 1/10 = 2
(20 × 101 )
= 22 = 4
Example 5·36. Find the geometric mean for the following distribution :
Marks
No. of students
:
:
0—10
5
10—20
7
20—30
15
30—40
25
40—50
8
Solution.
Marks
0—10
10—20
20—30
30—40
40—50
Mid-Point (X)
5
15
25
35
45
No. of Students (f)
5
7
15
25
8
log X
0·6990
1·1761
1·3979
1·5441
1·6532
N = 60
Geometric mean = Antilog
[
∑ f log X
N
] = Antilog [
84·5243
60
]
f. log X
3·4950
8·2327
20·9685
38·6025
13·2256
84·5243
= Antilog [1·40874] = 25·64 marks.
Example 5·37. The geometric mean of 10 observations on a certain variable was calculated as 16·2. It
was later discovered that one of the observations was wrongly recorded as 12·9; in fact it was 21·9. Apply
appropriate correction and calculate the correct geometric mean.
Solution. Geometric mean G of n observations is given by :
G = (X 1 X2 ,…, Xn )1/n
⇒
G n = X1 X2 …Xn
…(*)
Thus the product of the numbers is given by :
X1 X2 … Xn = G n = (16·2)10
[Given n = 10, G = 16·2]
…(**)
If the wrong observation 12·9 is replaced by the correct values 21·9, then the corrected value of the
product of 10 numbers is obtained on dividing the expression in (**) by wrong observation and multiplying
by the correct observation. Thus,
10
Corrected product (X1X2 …Xn ) = (16·2)12·9× 21·9
Hence, corrected value of the geometric mean G′, (say), is given by :
G′ =
⇒
⇒
[
(16·2)10 × 21·9
12·9
]
1/10
[
] [
]
1
= 10
[ 10 × 1·2095 + 1·3404 – 1·1106 ] = 101 [ 12·0950 + 1·3404 – 1·1106 ] = 12·3248
10 = 1·2325
1
1
log G′ = 10 log (16·2)10 + log 21·9 – log 12·9 = 10
10 log 16·2 + log 21·9 – log 12·9
G′ = Antilog (1·2325) = 17·08
Aliter. Using (5·26b) directly, we get : G′ = G .
( )
X1 ′
X1
1/10
= 16.2 ×
( )
21·9
12·9
1/10
Example 5·38. Three groups of observations contain 8, 7 and 5 observations. Their geometric means
are 8·52, 10·12 and 7·75 respectively. Find the geometric mean of the 20 observations in the single group
formed by pooling the three groups.
Solution. In the usual notations, we are given that :
n1 = 8, n2 = 7, n3 = 5; G1 = 8·52, G2 = 10·12, G3 = 7·75
The geometric mean G of the combined group of size N = n1 + n2 + n3 = 8 + 7 + 5 = 20, is given by :
1
[
]
log G = N n1 log G1 + n2 log G2 + n3 log G3 =
1
20
[ 8 log 8·52 + 7 log 10·12 + 5 log 7·75 ]
5·54
BUSINESS STATISTICS
[
]
[
]
1
1
= 20
8 × 0·9304 + 7 × 1·0052 + 5 × 0·8893 = 20
7·4432 + 7·0364 + 4·4465 = 18·9261
20 = 0·9463
∴
G = Antilog (0·9463) = 8·837
Example 5·39. Find the missing information in the following table :
Groups
B
8
—
7
A
10
20
10
C
—
6
—
Combined
Number
24
Mean
15
Geometric Mean
8·397.
[Delhi Univ. B.Com, (Hons.), 1998]
Solution. Taking the groups A, B and C as groups 1, 2 and 3 respectively, in the usual notations, we are
given :
Group A
Group B
Group C
Combined
Number
: n1 = 10
n2 = 8
n3 = ?
n1 + n2 + n 3 = 24
…(i)
–
–
–
–
Mean
: x 1 = 20
x2 = ?
x3 = 6
x = 15
…(ii)
Geometric Mean : G1 = 10
G2 = 7
G3 = ?
G = 8·397 …(iii)
From (i), we get
n3 = 24 – n1 – n2 = 24 – 10 – 8 = 6
…(iv)
–
–
–
–
–x = n1 x 1 + n2 x 2 + n3x3 = 10 × 20 + 8x 2 + 6 × 6 = 15 (Given)
n1 + n2 + n3
24
–x = 124 = 15·5
⇒
8 –x = 15 × 24 – 200 – 36 = 124
⇒
…(v)
2
n1 log G1 + n2 log G2 + n3 log G3
⇒
n1 + n2 + n3
10 × 1 + 8 × 0·8451 + 6 log G3 = 24 × 0·9242
log G
∴
⇒
2
=
G3 = Antilog
(
22·1808 – 10 – 6·7608
6
)
= Antilog
8
10 log 10 + 8 log 7 + 6 log G3
24
( )
5·42
6
= log (8·397)
= Antilog (0·9033) = 8·004 ~– 8 (approx.).
Example 5·40. Find the average rate of increase in population which in the first decade had increased
by 20%, in the next by 30% and in the third by 40%.
Solution. Since we are dealing with rate of increase in population, the appropriate average to be
computed is the geometric mean and not the arithmetic mean.
( n1 ∑ log X )
= Antilog ( 6·3392
3 ) = Antilog (2·1131)
G.M. = Antilog
= 129·7
CALCULATIONS FOR GEOMETRIC MEAN
log X
Decade Rate of growth Population at the end
of population
of the decade (X)
————————————————————————————————————
———————————————————————————————————
—————————————
1
2
3
Total
20%
30%
40%
120
130
140
2·0792
2·1139
2·1461
∑ log X = 6·3392
Hence, the average percentage rate of increase in the population per decade over the entire period is :
129·7 – 100 = 29·7.
Example 5·41. Under what condition is geometric mean indeterminate ?
If the price of a commodity doubles in a period of 4 years, what is the average annual percentage
increase ?
Solution. Geometric mean is indeterminate if any one of the given observations is negative. In this
case G.M. becomes imaginary. In general, G.M. is indeterminate (imaginary) if odd number of given
observations are negative.
5·55
AVERAGES OR MEASURES OF CENTRAL TENDENCY
Also, if any one of the given observations is zero, G.M. becomes zero irrespective of the size of the
other observations.
In the usual notations we are given :
P0 = Rs. x, (say) ; Pn = Rs. 2x ; n = 4.
If r is the average annual percentage increase in price over this period then we have :
Pn = P0 (1 + r) n ⇒
2x = x (1 + r)4
(1 + r)4 = 2
⇒
(1 + r) = 21/4
⇒
(
1 + r = Antilog
1
4
)
log 2 = Antilog
(
0·3010
4
) = Antilog (0·07525) = 1·190
∴
r = 1·190 – 1 = 0·19
Hence, the average annual percentage increase is 0·19 i.e., 19%.
Example 5·42. An assessee depreciated the machinery of his factory by 10% each in the first two years
and by 40% in the third year and thereby claimed 21% average depreciation relief from taxation
department, but the I.T.O. objected and allowed only 20%. Show which of the two is right.
Solution.
COMPUTATION OF A.M. AND G.M.
The average value (arithmetic mean) at the
end of three years is :
–
X=
∑X 240
3 = 3
Year
Rate of
Value at the end
depreciation of the year (X)
log X
——————————————————————————————
———————————————————————————
—————————————————————————
1
2
3
= 80
10%
10%
40%
90
90
60
1·9542
1·9542
1·7782
Hence, the average rate of depreciation per
annum for the entire period of 3 years is
100 – 80 = 20%.
∑ X = 240
∑ log X = 5·6866
This can be computed otherwise by taking the average of 10%, 10%, 40% which is :
————————————
——————————————————————————————————————————
——————————————————————————
10 + 10 + 40
3
The geometric mean is given by :
G.M. = Antilog
(
1
3
)
= 60
3 = 20%
∑log X = Antilog
(
5·6866
3
) = Antilog (1·8955) = 78·61
Hence, the average (geometric mean) rate of depreciation per annum for the entire period of three years
is 100 – 78·61 = 21·39% ~– 21%.
The assessee had claimed 21% depreciation using G.M. while the I.T.O. objected and allowed 20%
depreciation using A.M.
Since we are dealing with rates, the arithmetic mean does not depict the average depreciation correctly.
Geometric mean is the correct average to be used. Hence, I.T.O. was wrong in not allowing 21%
depreciation as claimed by the assesse.
Example 5·43. (a) Show that in finding the A.M. of a set of readings on a thermometer, it does not
matter whether we measure the temperature in Centigrade (C) or Fahrenheit (F) degrees.
[Delhi Univ. B.A. (Econ. Hons.) 2000]
(b) However, in computing Geometric Mean, it does matter which readings we use.
Solution. If F and C be the readings in Fahrenheit and Centigrade respectively then we have the
relation :
F – 32 C
9
=
⇒
F = 32 + C
180
100
5
Thus the Fahrenheit equivalents of C1 , C 2 , … , Cn are
32 + 59 C 1 , 32 + 95 C 2 , …, 32 + 95 C n respectively.
5·56
BUSINESS STATISTICS
Hence, the arithmetic mean of the readings in Fahrenheit is
= 1n
{(
{
) (
)
(
32 + 59 C 1 + 32 + 59 C 2 + … + 32 + 59 Cn
}
= 1n 32n + 59 (C1 + C2 + … + Cn ) = 32 + 95
(
)}
C1 + C2 + … + Cn
n
9 –
)
–
= 32 + 5 C , which is the Fahrenheit equivalent of C .
Hence, in finding the arithmetic mean of a set of n readings on a thermometer, it is immaterial whether
we measure temperature in Centigrade or Fahrenheit.
Geometric mean G, of n readings in Centigrade is : G = (C 1 C2…C n )1/n
Geometric mean G1, (say), of Fahrenheit equivalents of C 1 , C 2 ,…,C n is
{ ( 32 + C ) ( 32 + C ) … ( 32 + C ) }
which is not equal to the Fahrenheit equivalent of G viz., { ( C . C …C ) + 32 } .
9
5
G1 =
1
9
5
1/n
9
5
2
9
5
n
1/n
1
2
n
Hence, in finding the geometric mean of the n readings on a thermometer, the scale (Centigrade or
Fahrenheit) is important.
Example 5.44. If the arithmetic mean of two unequal positive real number ‘a’ and ‘b’, (a > b), be
twice as great as their geometric mean, show that
a : b = (2 + ⎯
√ 3) : (2 – √
⎯ 3)
[I.C.W.A. (Foundation), June 2001]
Solution. The arithmetic mean (A.M.) and the geometric mean (G.M.) of two unequal positive real
numbers a and b, (a > b), are given by :
a+b
A.M. =
and
G.M. = ⎯
√⎯ab .
2
We are given :
a+b
A.M. = 2 G.M.
⇒
=2⎯
√⎯ab
⇒
a+b=4⎯
√⎯ab
…(i)
2
Also (a – b) 2 = (a + b)2 – 4ab = 16 ab – 4 ab = 12 ab
[From (i)]
⇒
(a – b) = ± 2 √
⎯ 3 ⎯√⎯ab
⇒
Adding and subtracting (i) and (ii), we get respectively :
a–b=2⎯
√ 3 ⎯√⎯ab (Q a > b)
2a = 2 √
⎯⎯ab (2 + √
⎯ 3) …(iii)
and
2b = 2 ⎯
√⎯ab (2 – ⎯√ 3)
Dividing (iii) by (iv), we get :
a 2+√
⎯3
=
⇒
a : b = (2 + ⎯
√ 3) : (2 – ⎯√ 3)
b
2 –√
⎯3
…(ii)
…(iv)
5·9·5. Weighted Geometric Mean. If the different values X1 , X2,…, Xn of the variable are not of equal
importance and are assigned different weights, say, W1, W 2 ,…,Wn respectively according to their degree of
importance then their weighted geometric mean G.M. (W) is given by
G.M.(W) = (X 1 W1 × X2 W2 × … × Xn Wn )1/N
…(5·27)
where
N = W1 + W2 + … + W n = ∑W, is the sum of weights.
Taking logarithm of both sides in (5·27), we get
[
]
1
log [G.M. (W)] = 1N W1 log X1 + W2 log X2 + … + W n log Xn = N
∑W log X
⇒
G.M. (W) = Antilog
[ 1N ∑
]
W log X = Antilog
[
∑W log X
∑W
]
…(5·27a)
5·57
AVERAGES OR MEASURES OF CENTRAL TENDENCY
Example 5·45. The weighted geometric mean of the four numbers 8, 25, 19 and 28 is 22·15. If the
weights of the first three numbers are 3, 5, 7 respectively, find the weight (positive integer) of the fourth
number.
Solution. Let the weight of the fourth number be w.
Weighted Geometric Mean (G) = 22·15 (Given)
Also
log X
log G = ∑W∑W
log X
log 22·15 = ∑W∑W
⇒
COMPUTATION OF WEIGHTED G.M.
⇒
log (22·15)
⇒
⇒
⇒
⇒
⇒
(15 + w) × 1·3454
15 × 1·3454 + 1·3454w
20·1810 + 1·3454w
1·4472w – 1·3454w
0·1018w
⇒
w
+ 1·4472w
= 18·6504
15 + w
= 18·6504 + 1·4472w
= 18·6504 + 1·4472 w
= 18·6504 + 1·4472w
= 20·1810 – 18·6504
= 1·5306
= 1·5306
1·1018 =
X
W
log X
W log X
———————————————
———————————————
————————————————
—————————————————————————
8
25
19
28
0·9031
1·3979
1·2788
1·4472
3
5
7
w
2·7093
6·9895
8·9516
1·4472w
————————————————
———————————————
———————————————
——————————————————————————
Total
15 + w
18·6504 + 1·4472w
15 approx.
5·10. HARMONIC MEAN
If X 1 , X 2 ,…, Xn is a given set of n observations, then their harmonic mean, abbreviated as H.M. or
simply H is given by :
1
1
H =
=
= n1
…(5·28)
1 1
1
1
1
1
∑
+
+…+
∑
X
n X1 X2
Xn
n
X
]
[
()
( )
In other words, Harmonic Mean is the reciprocal of the arithmetic mean of the reciprocals of the given
observations.
In case of frequency distribution, we have
f1 f2
fn
f
1 1
1
N
=
+
+…+
= ∑
⇒
H=
…(5·28a)
H N X1 X2
Xn
N
X
∑ (f/X)
[
]
( )
where N = ∑ f, is the total frequency, X is the value of the variable or the mid-value of the class (in case of
grouped or continuous frequency distribution) and f is the corresponding frequency of X.
Remark. Effect of Change of Scale on Harmonic Mean. Let x1 , x2, …, xn be the given set of
n observations. If the variable x is transformed to the new variable u by change of scale : u = kx, k ≠ 0 …(*)
then, by definition
[
] [
]
1
1 1
1
1
1 1
1
1
=
+ +…+
=
+
+…+
H.M. (u) n u1 u2
un
n kx1 kx2
kxn
[
]
[Using (*)]
1 1 1 1
1
1
1
= ·
+ +…+
= ·
k n x1 x2
xn
k H.M. (x)
⇒
H.M.(u) = k × H.M. (x)
Hence, we have the following result :
If u = kx, k ≠ 0, then
H.M. (u) = k × H.M. (x)
5·10·1. Merits and Demerits of Harmonic Mean.
Merits : (i) Harmonic mean is rigidly defined.
(ii) It is based on all the observations.
(iii) It is suitable for further mathematical treatment.
…[5·28b]
5·58
BUSINESS STATISTICS
If H 1 and H2 are the harmonic means of two groups of sizes N1 and N 2 respectively, then the harmonic
mean H of the combined group of size N 1 + N2 is given by :
[
N1
N2
1
1
=
+
H
N 1 + N 2 H1
H2
]
…(5·28c)
(iv) Since the reciprocals of the values of the variable are involved, it gives greater weightage to
smaller observations and as such is not very much affected by one or two big observations.
(v) It is not affected very much by fluctuations of sampling.
(vi) It is particularly useful in averaging special types of rates and ratios where time factor is variable
and the act being performed remains constant.
Demerits. (i) It is not easy to understand and calculate.
(ii) Its value cannot be obtained if any one of the observations is zero.
(iii) It is not a representative figure of the distribution unless the phenomenon requires greater
weightage to be given to smaller items. As such, it is hardly used in business problems.
Uses. As has been pointed out in merit (vi) , harmonic mean is specially useful in averaging rates and
ratios where time factor is variable and the act being performed e.g., distance is constant. The following
examples will clarify the point.
Example 5·46. The following table gives the weights of 31 persons in a sample enquiry. Calculate the
mean weight using (i) Geometric mean and (ii) Harmonic mean.
Weight (lbs.) :
No. of persons :
130
3
135
4
140
6
145
6
146
3
148
5
149
2
150
1
157
1
Solution.
COMPUTATION OF G.M. AND H.M.
Weight (lbs.)
(X)
No. of persons
(f)
log X
f log X
1
X
f
X
130
135
140
145
146
148
149
150
157
3
4
6
6
3
5
2
1
1
2·1139
2·1303
2·1461
2·1614
2·1644
2·1703
2·1732
2·1761
2·1959
6·3417
8·5212
12·8766
12·9684
6·4932
10·8515
4·3464
2·1761
2·1959
0·00769
0·00741
0·00714
0·00690
0·00685
0·00676
0·00671
0·00667
0·00637
0·02307
0·02964
0·04284
0·04140
0·02055
0·03380
0·01342
0·00667
0·00637
∑f = N = 31
G.M. = Antilog
H.M. =
(
∑f log X = 66·7710
1
N ∑f
log X
) = Antilog (
66·7710
31
∑(f / X) = 0·21776
) = Antilog (2·1539) = 142·5
N
31
=
= 142·36
∑(f /X) 0·21776
Hence, the mean weight of 31 persons using (i) geometric mean is 142·5 lbs. and (ii) harmonic mean is
142·36 lbs.
Example 5.47. If 2u = 5x, is the relation beween two variables u and x, and harmonic mean of x is
0·4, find the harmonic mean of u.
[I.C.W.A. (Foundation), June 2005]
Solution.
2u = 5x
⇒
5
u= x
2
…(*)
5·59
AVERAGES OR MEASURES OF CENTRAL TENDENCY
Also
H.M. (x) = 0.4
(Given)
…(**)
If u 1 , u 2 , …, un are n observations corresponding to x1, x 2 , …, xn respectively, obtained by the
transformation (*) so that
ui =
5
x , (i = 1, 2, …, n),
2 i
…(***)
then, by definition
H.M. (u) =
1
[
1 1
1
1
+ +…+
n u1 u2
un
=
1
=
] [
5 ⎡
1
2 ⎢1 1 1
1
⎢
+ +…+
n
x
x
x
1
2
n
⎢⎣
[
1 21 2 1 … 21
+ · +
+
n 5 x1 5 x2
5 xn
]
[From (***)]
]
⎤ = 5 H.M. (x) = 5 × 0·4 = 1
⎥ 2
2
⎥
⎥⎦
[From (**)]
Remark. We may state and use the result (5·28b) directly, to get from (*)
H.M. (u) =
5
5
H.M. (x) = × 0.4 = 1
2
2
Example 5.48. Find the harmonic mean of
1, 2, 3, … , n respectively.
Solution. The harmonic mean (H ) is
given by :
1 ∑ (f/x)
=
H
∑f
2 + 3 + 4 + … + n + (n + 1)
=
1+2+3+…+n
(1 + 2 + 3 + … + n) + n
=
(1 + 2 + 3 + … + n)
n
=1+
1+2+3+…+n
=1+
∴
1
2
3
n
,
,
, … ,
occurring with frequencies
2
3
4
n+1
[I.C.W.A. (Foundation), June 2006]
x
f
(1)
(2)
1/2
2/3
3/4
M
(n – 1)/n
n/(n + 1)
1
2
3
M
n–1
n
1
x
(3)
f
x
(4) = (2) × (3)
2
3/2
4/3
2
3
4
n/(n – 1)
(n + 1)/n
n
n+1
n
2
n+3
=1+
=
n (n + 1)/2
n+1 n+1
H = (n + 1)/(n + 3)
Example 5·49. A cyclist pedals from his house to his college at a speed of 10 km. p.h. and back from
the college to his house at 15 km. p.h. Find the average speed.
Solution. Let the distance from the house to the college be x kms.
In going from house to college, the distance (x kms) is covered in x/10 hours, while in coming from
college to house, the distance is covered in x / 15 hours. Thus a total distance of 2x kms is covered in
(
x
x
10 + 15
) hours.
2x
2
=
= 12 km. p.h.
x
x
1
1
+
+
10 15
10 15
Remarks. 1. In this case the average speed is given by the harmonic mean of 10 and 15 and not by the
arithmetic mean.
Hence, average speed =
Total distance travelled
=
Total time taken
(
) (
)
5·60
BUSINESS STATISTICS
2. If equal distances are covered (travelled) per unit of time with speeds equal to V1 , V2 ,…,Vn , say, then
the average speed is given by the harmonic mean of V1 , V2 ,…, Vn i.e.,
n
Average speed =
= n
1
1
1
+
+…+
∑ 1V
V1
V2
Vn
Example 5·50. A vehicle when climbing up a gradient, consumes petrol at the rate of 1 litre per 8 kms.
while coming down it gives 12 kms. per litre. Find its average consumption for to and fro travel between
two places situated at the two ends of a 25 km. long gradient. Verify your answer.
[Delhi Univ. B.A. (Econ. Hons.), 1994]
Solution. Since the consumption of petrol is different for upward and downward journeys (at a
constant distance of 25 km.), the appropriate average consumption for to and fro journey is given by the
harmonic mean of 8 km. and 12 km.
∴
Average consumption for to and fro journey
1
2
48
=
=
=
= 9·6 km. per litre
…(*)
1 1 1
3+2
5
+
2 8 12
24
Verification. Consumption of petrol for upward journey of 25 km. @ 1 litre per 8 kms. = 25
8 litres.
(
) ()
(
) (
)
25
Consumption of petrol for downward journey of 25 km @ 1 litre per 12 kms. = 12
litres.
∴
Total petrol consumed for total journey of 25 + 25 = 50 km. is
25 25
1 1
25
125
+
litres = 25
+
litres =
(3 + 2) litres =
litres
8 12
8 12
24
24
∴
Average consumption for to and fro journey
Total distance
50 × 24 48
=
=
=
= 9·6 km. per litre, which is same as in (*).
Total petrol used
125
5
Example 5·51. In a certain office, a letter is typed by A in 4 minutes. The same letter is typed by B, C
and D in 5, 6, 10 minutes respectively. What is the average time taken in completing one letter ? How many
letters do you expect to be typed in one day comprising of 8 working hours ?
[Delhi Univ. B.A. (Econ. Hons.), 1996; 1995]
Solution. The average time (in minutes) taken by each of A, B, C and D in completing one letter is the
harmonic mean of 4, 5, 6 and 10 given by :
4
4
240
=
=
= 5·5814 minutes per letter.
1 1 1 1
43
15 + 12 + 10 + 6
+ + +
4 5 6 10
60
43
Hence, expected number of letters typed by each of A, B, C and D is
letters per minute.
240
Hence, in a day comprising of 8 hours = 8 × 60 minutes, the total number of letters typed by all of
them (A, B, C and D) is :
43
× 4 × 8 × 60 = 344.
240
Example 5·52. An investor buys Rs. 1,200 worth of shares in a company each month. During the first 5
months he bought the shares at a price of Rs. 10, Rs. 12, Rs. 15, Rs. 20, and Rs. 24 per share. After 5
months what is the average price paid for the shares by him ?
Solution. Since the share value is changing after a fixed unit time (1 month), the required average price
per share is the harmonic mean of 10, 12, 15, 20 and 24 and is given by :
5
5
5 × 120
=
=
= Rs. 14·63.
41
1
1
1
1
1
12 + 10 + 8 + 6 + 5
+
+
+
+
10 12 15 20 24
120
(
)
(
)
(
(
)
) (
Note. For an alternate solution, see Example 5·8.
)
5·61
AVERAGES OR MEASURES OF CENTRAL TENDENCY
5·10·2. Weighted Harmonic Mean. Instead of fixed (constant) distance being travelled with varying
speeds (c.f. Remark 2, Example 5·49), let us now suppose that different distances are travelled with
corresponding different speeds. In that case what is going to be the average speed ?
Let us suppose that distances s1 , s 2 ,…, sn , are travelled with speed v 1 , v 2 ,…, vn per unit of time. If
t1 ,t2,…, tn are the respective times taken to cover these distances then we have :
s1
s2
sn
t1 = ,
t2 = ,
…,
tn =
…(*)
v1
v2
vn
Total distance travelled s1 + s2 + … + sn
∴
Average speed =
=t +t +…+t
Total time taken
1
2
n
s1 + s2 + … + sn
= s s
[From (*)]
sn
1
2
+ +…+
v1 v2
vn
[
=
]
∑s
∑
1
=
…(5·30)
( ) ( )∑( )
s
v
1
∑s
s
v
which is the weighted harmonic mean of the speeds, the corresponding weights being the distances covered.
Hence, if different distances are travelled with corresponding different speeds, then the average speed
is given by the weighted harmonic mean of the speeds, the corresponding weights beings the distances
covered.
Example 5·53. You make a trip which entails travelling 900 kms. by train at an average speed of
60 km. p.h.; 3000 kms. by boat at an average of 25 km. p.h.; 400 kms. by plane at 350 km. p.h., and finally,
15 kms by taxi at 25 km. p.h. What is your average speed for the entire distance ?
Solution. Since different distances are covered with
varying speeds, the required average speed is given by the
weighted harmonic mean of the speeds (in km. p.h) 60, 25, 350
and 25; the corresponding weights being the distances covered
(in kms.) viz., 900, 3000, 400 and 15 respectively.
∴
∑W
4315
Average Speed = ∑(W/X)
= 137·03
= 31·49 km. p.h.
COMPUTATION OF WEIGHTED H.M.
X
W
W/X
——————————————
———————————————————
———————————————————————————
60
25
350
25
900
3000
400
15
15
120
1·43
0·60
—————————————————
————————————————————
—————————————————————————
∑W = 4315
∑(W/X) = 137·03
5·11. RELATION BETWEEN ARITHMETIC MEAN, GEOMETRIC MEAN AND HARMONIC
MEAN
The arithmetic mean (A.M.), the geometric mean (G.M.) and the harmonic mean (H.M.) of a series of
n observations are connected by the relation :
A.M. ≥ G.M. ≥ H.M.
…(5·31)
the sign of equality holding if and only if all the n observations are equal.
Remark. For two numbers we also have
G2 = A × H
…(5·32)
where A, G and H represent arithmetic mean, geometric mean and harmonic mean respectively.
Proof : Let a and b be two real positive numbers i.e., a > 0, b > 0.
a+b
1
2ab
Then
A.M. =
;
G.M. = √
⎯⎯ab
;
H.M. =
=
…(*)
2
a+b
1 1 1
+
2 a b
(
∴
A×H=
a + b 2ab
·
= ab = G2
2
a +b
)
5·62
BUSINESS STATISTICS
Example 5·54. H.M., A.M. and G.M. of a set of 5 observations are 10·2, 16 and 14 respectively.”
Comment.
Solution. We are given : n = 5; A.M. = 16 ; G.M. = 14 and H.M. = 10·2. Since A.M. > G.M. > H.M.,
the above statement is correct.
Example 5·55. The arithmetic mean of two observations is 127·5 and their geometric mean is 60. Find
(i) their harmonic mean and (ii) the two observations.
[Delhi Univ. B.Com. (Hons.), (External), 2007]
Solution. (i) Let the two observations be a and b. Then we are given :
a+b
Arithmetic Mean =
= 127·5
⇒
a + b = 255
…(*)
2
G.M. = ⎯
√⎯⎯
a × b = 60
⇒
ab = 3600
Harmonic mean of two numbers a and b is given by :
2ab 2 × 3600 480
H.M. =
=
=
= 28·24
a+b
255
17
Aliter. For two numbers, we have :
2
(ii) We have
(a – b) 2 = (a + b)2 – 4ab = (255)2 – 4 × 3600
= 65025 – 14400 = 50625
a – b = ± ⎯√⎯⎯⎯
50625 = ± 225
a + b = 255 and a – b = 225
Adding, we get
[From (*) and (**)]
2
60
480
H=G
A = 127·5 = 1 7 = 28·24
⇒
G2 = AH
…(**)
[From (*) and (**)]
∴
2a = 480
∴
⇒
b = 255 – a
= 255 – 240 = 15
a + b = 225 and a – b = – 225
Adding, we get
a = 480
2 = 240
[From (*)]
2a = 30
∴
⇒
b = 255 – a
= 255 – 15 = 240
a = 30
2 = 15
[From (*)]
Hence, the two observations are 240 and 15
5·12. SELECTION OF AN AVERAGE
From the discussion of the merits and demerits of the various measures of central tendency in the
preceding sections, it is obvious that no single average is suitable for all practical problems. Each of the
averages has its own merits and demerits and consequently its own field of importance and utility. For
example, arithmetic mean is not to be recommended while dealing with frequency distribution with extreme
observations or open end classes. Median and mode are the averages to be used while dealing with open
end classes. In case of qualitative data which cannot be measured quantitatively (e.g., for finding average
intelligence, honesty, beauty, etc.), median is the only average to be used. Mode is particularly used in
business and geometric mean is to be used while dealing with rates and ratios. Harmonic mean is to be used
in computing special types of average rates or ratios where time factor is variable and the act being
performed e.g., distance, is constant.
Hence, the averages cannot be used indiscriminately. For sound statistical analysis, a judicious
selection of the average depends upon :
(i) the nature and availability of the data,
(ii) the nature of the variable involved,
(iii) the purpose of the enquiry.
AVERAGES OR MEASURES OF CENTRAL TENDENCY
5·63
(iv) the system of classification adopted, and
(v) the use of the average for further statistical computations required for the enquiry in mind.
However, since arithmetic mean :
(i) satisfies almost all the properties of an ideal average as laid down by Prof. Yule,
(ii) is quite familiar to a layman, and
(iii) has very wide applications in statistical theory at large,
it may be regarded as the best of all the averages.
5·13. LIMITATIONS OF AVERAGES
In spite of its very wide applications in statistical analysis, the averages have the following limitations :
1. Since average is a single numerical figure representing the characteristics of a given distribution,
proper care should be taken in interpreting its value otherwise it might lead to very misleading conclusions.
In this context, it might be appropriate to quote a classical joke regarding average about a village school
teacher who had to cross a river along with his family. On enquiry he was given to understand that the
average depth of the river was 3 feet. He measured the heights of the members of the family (himself, his
wife, 2 daughters and 3 sons) and found that their average (mean) height was 321 feet. Since the average
height of the family came out to be higher than the average depth of the river, he ordered his family to cross
the river. But when he reached the other side of the river, three of his children were missing. He again
checked his arithmetical calculations which still gave the same result and was wondering as to what and
where was the mistake. He wrote a couplet in Urdu, reading :
‘Arba jyon ka tyon
Kunba dooba kyon’
(Arba means arithmetic or calculations and Kunba means family). In fact, the teacher had the
misconception about the average depth of the river which he mistook for uniform depth but in fact the river
was very shallow in the beginning but became deeper and deeper and in the middle it was as deep as 4 feet
or so. Accordingly, the members of the family with height bellow 4 feet were drowned.
2. A proper and judicious choice of an average for a particular problem is very important. A wrong
choice of the average might give wrong and fallacious conclusions.
3. An average fails to give the complete picture of a distribution. We might come across a number of
distributions having the same average but differing widely in their structure and constitution. To form a
complete idea about the distribution, the measures of central tendency are to be supplemented by some
more measures such as dispersion, skewness and kurtosis.
4. In certain types of distributions like U-shaped or J-shaped distributions, an average (which is only a
single point of concentration) fails to represent the entire series [c.f. Chapter 4].
5. Sometimes an average might give very absurd results. For instance, the average of a family might
come out in fractions which is obviously absurd. In this context we might quote the following :
“The figure of 2·2 children per adult female is felt in some respects to be absurd and the Royal
Commission suggested that the middle classes be paid money to increase the average to a rounder and more
convenient number”.
EXERCISE 5·4
1. Define Geometric Mean and discuss its merits and demerits. Give two practical situations where you will
recommend its use.
2. (a) “It is said that the choice of an average depends on the particular problem in hand”.
Examine the above statement and give at least one instance each for the use of Mode and Geometric Mean.
(b) Discuss the strong and weak points of various measures of central tendency.
3. (a) What are the advantages and disadvantages of the chief averages used in Statistics ? Indicate their special
uses if any.
(b) What are the desideratta for a satisfactory average ? Examine the geometric mean in the light of these
desideratta and bring out the special properties of this average which lend to its use in intercensal population counts and
in the construction of index numbers.
5·64
BUSINESS STATISTICS
4. (a) “Each average has its own special features and it is difficult to say which one is the best”. Explain and
illustrate.
(b) Why is arithmetic mean generally preferred over median as the measure of central tendency ? What is the
relation between arithmetic mean and geometric mean ? When is the latter preferred over the former ?
5. Explain the relative merits of geometric mean over other measures of central tendency.
6. Give a specific example of an instance in which :
(a) The median would be used in preference to arithmetic mean,
(b) The arithmetic mean would not be as satisfactory as the geometric mean, and
(c) Mode would be used in preference to the median.
1 1
7. (a) Find the G.M. of 1, 2, 3, , · What will be the geometric mean if ‘0’ is added to this set of values ?
2 3
[I.C.W.A. (Foundation), June 2003]
1/5
1 1
Ans. G = 1 × 2 × 3 × ×
= 11/5 = 1 ; G = 0
2 3
(
)
(b) Find the geometric mean of : 1, 7, 18, 65, 91 and 103.
Ans. 20·62.
(c) Calculate geometric mean of the data : 1, 7, 29, 92, 115 and 375.
Ans. 30·50
8. Calculate arithmetic mean and geometric mean of the following distribution
x
:
2
3
4
5
6
7
8
f
:
2
4
6
2
3
2
1
Ans. A.M. = 4·5 ; G.M. 4·192.
9. If the population has doubled itself in twenty years, is it correct to say that the rate of growth has been 5% per
annum ?
Ans. No. r = 3·5%.
10. The population of a city was 1,00,000 in 1975 and 1,44,000 a decade later. Estimate the population at the
middle of the decade.
[Delhi Univ. B.A. (Econ. Hons.) 1996]
Hint and Ans. r : Percentage rate of growth per annum.
Then
1,44,000 = 1,00,000 (1 + r)10
⇒
(1 + r)10 = 144
=
100
( )
12
10
2
⇒
(1 + r)5 = 12
= 1.2.
10
Estimated population at the middle of the decade = 1,00,000 (1 + r)5 = 1,00,000 × 1·2 = 1,20,000.
11. The population of India in 1951 and 1961 were 361 and 439 million respectively.
(i) What was the average percentage increase per year during the period ?
(ii) If the average rate of increase from 1961 to 1971 remains the same, what would be the population in 1971 ?
Ans. (i) 2%, (ii) 533·85 million.
12. The population of a country increased by 20 per cent in the first decade and by 30 per cent in the second
decade and by 45 per cent in the third decade. Determine the average decennial growth rate of population.
Ans. 31·3%
13. A machine depreciates by 40% in the first year, by 25% in the second year and by 10% per annum for the next
three years, each percentage being calculated on the diminishing value. What is the average percentage of depreciation
for the entire period ?
[Delhi Univ. B.Com. (Hons.), 1994]
Ans. 20%.
14. (a) An income-tax assessee depreciated the machinery of his factory by 20 per cent in each of the first two
years and 40 per cent in the third year. How much average depreciation relief should he claim from the taxation
department ?
[Delhi Univ. B.A. (Econ. Hons.), 1999; Bangalore Univ. B.Com., 1998]
Ans. 27·32%.
(b) A businessman depreciated the machinery of his factory by 20% in the first two years and 40% in the third
year. What is the average depreciation for the three years ?
[Delhi Univ. B.A. (Econ. Hons.), 2004]
Ans. G.M. = 27.32%.
15. (a) An economy grows at the rate of 2% in the first year, 2·5% in the second year, 3% in the third, 4% in the
fourth,…and 10% in the tenth year. What is the average rate of growth of the economy ?
Ans. 5·6% p.a.
(Nagarjuna Univ. B.Com., 1996)
AVERAGES OR MEASURES OF CENTRAL TENDENCY
5·65
(b) The annual rates of growth achieved by a nation for 5 years are 5%, 7·5%, 2·5%, 5% and 10% respectively.
What in the compound rate of growth for the 5 year period ?
[Delhi Univ. B.A. (Eco. Hons.), 1993]
Ans. 5·9%.
16. The number of divorces per 1,000 marriages in a big city in India increased from 96 in 1980 to 120 in 1990.
Find the annual rate of increase of the divorce rate for the period 1980 to 1990.
[Delhi Univ. B.Com. (Hons.), 1994]
Hint. 120 = 96
(
r
1 + 100
)
10
r
1 + 100
=
⇒
( )
120
96
1/10
Ans. r = 2·26%.
17. If arithmetic mean and geometric mean of two values are 10 and 8 respectively, find the values.
Ans. 16, 4.
18. A man gets three successive annual rises in salary of 20%, 30% and 25% respectively, each percentage being
reckoned in his salary at the end of the previous year. How much better or worse would he have been if he had been
given three annual rises of 25% each, reckoned in the same way.
[Delhi Univ. B.A. (Econ. Hons.), 2006]
Ans. The man would be better in the second case by 0·31% of his starting salary in the 1st year.
19. The geometric mean of 4 items is 100 and of another 8 items is 3·162. Find the geometric mean of the 12
items.
Ans. 10.
20. (a) Geometric mean of n observations is found to be G. How will you find the correct value of the Geometric
Mean if some of the values used in its calculation are found to be wrong and should be replaced by correct values ?
(b) Geometric mean of 2 numbers is 15. If by mistake one figure is taken as 5, instead of 3, find correct geometric
mean.
[Delhi Univ. B.Com. (Hons.) 1993]
⎯√⎯ab
= 15 ⇒ ab = 225
225 225
=
= 45
Wrong observation, (say), a = 5
∴ b =
5
a
Correct value of a = 3 (Given)
Hint. Let the two numbers be a and b.
⎯√⎯⎯
∴
Correct G.M. = 3 × 45 = 11·62 Or Use Formula (5·26b)
(c) The geometric mean of four values was calculated as 16. It was later discovered that one of the value was
recorded wrongly as 32 when, in fact, it was 162. Calculate the correct geometric mean.
[Delhi Univ. B.Com. (Hons.), 2004]
Ans. Correct G.M. = (16) ×
( )
162
32
1/4
= 16 ×
3
= 24
2
[Using (5·26b)]
21. (a) Define simple and weighted geometric mean of a given distribution.
The weighted geometric mean of three numbers 229, 275 and 125 is 203. The weights for 1st and 2nd numbers are
2 and 4 respectively. Find the weight of the third.
Ans. 3.
(b) The weighted geometric mean of the four numbers 9, 25, 17 and 30 is 15·3. If the weights of the first three
numbers are 5, 3 and 4 respectively, find the weight of the fourth number.
Ans. 2 (approx.).
22. (a) Define Harmonic Mean and discuss its merits and demerits. Under what situations would you recommend
its use.
(b) Find the geometric and harmonic mean from the following data.
Items
:
1
2
3
4
5
6
7
8
9
10
Value
:
15
250
15·7
157
1·57
105·7
105
1·06
25·7
0·257
Ans. GM = 16·04; HM = 1·7637.
1 1 1 1
23. (a) Find the harmonic mean of the numbers
, , , , 1. [I.C.W.A. (Foundation), June 2004, Dec. 2002]
5 4 3 2
1 1
1
1
1
Hint.
= ∑
= ( 5 + 4 + 3 + 2 + 1)
⇒
H=
H 5
5
3
x
(b) If each of 3, 48 and 96 occurs once and 6 occurs thrice, verify that geometric mean is greater than harmonic
mean.
[I.C.W.A. (Foundation), Dec., June 2004]
()
5·66
BUSINESS STATISTICS
Ans. G.M. = 12 ; H.M. = 6.94 ; G.M. > H.M.
24. What do you mean by Weighted Harmonic Mean ? When will you use it instead of Simple Harmonic Mean ?
Explain by a practical situation.
25. It is said that “Choice of an average depends on the particular problem in hand.” Examine the statement and
give at least one instance each for the use of mode, geometric mean and harmonic mean. [Delhi B.Com. (Hons.), 2007]
26. From the following statements select any two which are correct and any three which are incorrect. In respect of
each of such statements selected by you, give your comments explaining briefly why you consider the statement correct
or incorrect :
(i) The median may be considered more typical than the mean because the median is not affected by the size of the
extremes ;
(ii) In a frequency distribution the true value of the mode cannot be calculated exactly ;
(iii) The Geometric Mean cannot be used in the averaging of index numbers because it gives undue importance to
small numbers ;
(iv) The Harmonic Mean of a series of fractions is the same as the reciprocal of the arithmetic mean of the series.
Ans. (i) T, (ii) T, (iii) F, (iv) F.
27. Show that the weighted harmonic mean of the first n natural numbers, where the weights are equal to the
corresponding numbers, is given by (n + 1)/2.
[Delhi Univ. B.A. (Econ. Hons.), 2003]
28. (a) An aeroplane flies around a square the sides of which measure 100 km. each. The aeroplane covers at a
speed of 100 km. per hour first side, at 200 km. per hour the second side, at 300 km. per hour the third side and at 400
km. per hour the fourth side. Use the correct mean to find the average speed around the square.
Ans. 192 km. p.h.
(b) Four factories emit a kilogram of pollutant each in 4, 5, 8 and 12 days respectively. What is the average rate of
pollutant discharge ? Use your answer to calculate the total pollutant discharged by the four factories in one week.
[Delhi Univ. B.A. (Econ. Hons.), 2005]
Hint. Find H.M. of 4, 5, 8, 12.
480
Ans. 1 kg. pollutant in
days per factory.
79
553
79
×4×7=
= 4.608 kg.
Total pollutant discharged by four factories per week =
120
480
29. (a) A railway train runs for 30 minutes at a speed of 40 miles an hour and then, because of repairs of the track
runs for 10 minutes at a speed of 8 miles an hour, after which it resumes its previous speed and runs for 20 minutes
except for a period of 2 minutes when it had to run over a bridge with a speed of 30 miles per hour. What is the average
speed ?
[Delhi Univ. B.Com. (Hons.), 2009]
Hint. Average Speed = (Total Distance covered) ÷ (Total time taken)
=
[(
)
]
8
40
30
40
× 30 +
× 10 +
× 18 +
× 2 ÷ (30 + 10 + 20) m.p.h.
60
60
60
60
Ans. 34·33 m.p.h.
(b) A train runs 25 miles at a speed of 30 m.p.h.; another 50 miles at a speed of 40 m.p.h.; then due to repairs of
the track runs for 6 minutes at a speed of 10 m.p.h. and finally covers the remaining distance of 24 miles at a speed of
24 m.p.h. What is the average speed in miles per hour ?
[Punjab Univ. B.Com., 1994]
Ans. 31·41 m.p.h.
30. (a) A cyclist covers his first three kilometres at an average speed of 8 kms. per hour, another 2 kms. at 9 kms.
per hour and the last 2 kms. at 4 kms. per hour. Find the average speed for the entire journey.
Ans. 6·38 kms. per hour.
(b) If X travels 8 km. at 4 km. per hour; 6 km. at 3 km. per hour and 4 km. at 2 km. per hour, what would be the
average rate per hour at which he travelled ?
[Delhi Univ. B.Com. (Pass), 1998]
Ans. Weighted H.M. = 3 km. p.h.
31. (a) A man travelled by car for 3 days. He covered 480 km. each day. On the first day he drove for 10 hours at
48 km. an hour on the second day he drove for 12 hours at 40 km. an hour and on last day he drove for 15 hours at
32 km. an hour. What was his average speed ?
[Bombay Univ. B.Com., 1996]
Ans. 38·919 km. p.h.
AVERAGES OR MEASURES OF CENTRAL TENDENCY
5·67
(b) Kishore travels 900 kms. by train at an average speed of 60 kms. per hour; 3,000 kms. by steamship at an
average of 25 kms. per hour; 400 kms. by aeroplane at 350 kms. per hour; and finally 15 kms. by bus at 25 kms. per
hour. Calculate his average speed for the entire journey.
[C.S. (Foundation), Dec. 2001]
Ans. 31·556 km. p.h.
32. A man travels from Agra to Dehradun covering 204 miles at a mileage rate of 10 miles per gallon of petrol and
via Ghaziabad with an additional journey of 40 miles at the rate of 15 miles per gallon. Find the average mileage per
gallon.
Ans. 10·58 miles per gallon.
33. The consumption of petrol by a motor was a gallon for 20 miles while going up from plains to hill station and a
gallon for 24 miles while coming down. What particular average would you consider appropriate for finding the
average consumption in miles per gallon for up and down journey, and why ?
Ans. Harmonic Mean = 21·82 m.p. gallon.
34. A man having to drive 90 kilometres wishes to achieve an average speed of 30 kilometres per hour. For the
first half of the journey he averages only 20 km. p.h. What must be his average for the second half of the journey if his
overall average is to be 30 km. p.h.
Ans. 60 km. p.h.
35. An aeroplane travels distances of d1, d2 and d3 kms. at speeds V1, V2 and V3 km. per hour respectively. Show
that the average speed (V) is given by :
d1 + d2 + d3
d1
d2
d3
= V + V + V
V
1
2
3
[Delhi Univ. B.A. (Econ. Hons.), 1993]
36. (a) A person purchases one kilogram of cabbage from each of the four places at the rate of 20 kg., 16 kg.,
12 kg. and 10 kg. per 100 rupee respectively. On the average how many kg. of cabbage has he purchased per 100
rupees ?
[Delhi Univ. B.Com. (Pass), 2001]
(b) If you spend Rs. 100 per week on apples and the price of apples for three weeks is Rs. 25, Rs. 20 and Rs. 10
per kilogram, what is the average price of apples for you ?
[Delhi Univ. B.A. (Econ. Hons.), 2002]
Ans. Rs. 15·79 per kg.
37. In a certain office a letter is typed by A in 4 minutes. The same letter is typed by B, C and D in 5, 6, 10 minutes
respectively. What is the average time taken in completing one letter ? How many letters do you expect to be typed in
one day comprising of 8 working hours.
480 ~
Ans. H.M. = 5·58 minutes per letter ; Letters typed in 8 hours (480 minutes) =
– 86.
5·58
38. A scooterist purchased petrol at the rate of Rs. 24, Rs. 29.50 and Rs. 36.85 per litre during three successive
years. Calculate the average price of petrol,
(i) If he purchased 150, 180 and 195 litres of petrol in the respective years and
(ii) If he spent Rs. 3,850, Rs. 4,675 and Rs. 5,825 in three years.
Give support to your answer.
[Delhi Univ. B.Com. (Hons.), 2005]
Total money spent on petrol
Hint. Average price of petrol/litre =
Total petrol consumed in litres
(i) Weighted A.M. of prices the weights being the quantities of petrol purchased
(ii) Weighted H.M. of prices, the weights being the money spent on petrol.
Ans. (i) Rs. 30·65/litre, (ii) Rs. 30/litre (approx.)
39. (a) Define Arithmetic Mean, Harmonic Mean and Geometric Mean for a set of n observations and state the
relationship between them.
Ans. A ≥ G ≥ H ; the sign of equality holds if and only if all the observations are equal.
(b) Show the relationship between arithmetic mean and harmonic mean for the variable X, which can take the
values a and b such that a, b are non-negative integers.
[Delhi Univ. B.A. (Econ. Hons.), 2007]
Ans. A × H =
( )( )
a+b
2ab
·
= ab = G2
2
a+b
(c) If for two numbers, the arithmetic mean is 25 and the harmonic mean is 9, what is the geometric mean of the
series ?
[C.A. (Foundation), May 2001]
Ans. G.M. = 15.
5·68
BUSINESS STATISTICS
(d) If A.M. of two numbers is 17 and G.M. is 15, find the H.M. of these numbers.
[Delhi Univ. B.A. (Econ. Hons.), 2000]
Ans. 13·24
40. (a) Comment on the following : “The G.M. and A.M. of a distribution are 27 and 30. Then H.M. is 26.”
[Delhi Univ. B.Com. (Hons.), 2009]
Ans. Since A.M. ≥ G.M. ≥ H.M., the statement is correct.
(b) State giving reasons which average will be more appropriate in the following cases :
(i) The distribution has open-end classes.
(ii) The distribution has wide range of variations.
(iii) When depreciation is charged by diminishing balance method and an average rate of depreciation is to be
calculated.
(iv) The distance covered is fixed but speeds are varying and an average speed is to be calculated.
[Delhi. Univ. B.Com. (Pass), 1999]
Ans. (i) Md or Mo, (ii) Md, (iii) G.M., (iv) H.M.
6
Measures of Dispersion
6·1. INTRODUCTION AND MEANING
Averages or the measures of central tendency give us an idea of the concentration of the observations
about the central part of the distribution. In spite of their great utility in statistical analysis, they have their
own limitations. If we are given only the average of a series of observations, we cannot form complete idea
about the distribution since there may exist a number of distributions whose averages are same but which
may differ widely from each other in a number of ways. The following example will illustrate this
viewpoint.
Let us consider the following three series A, B and C of 9 items each.
Series
Total Mean
A
15,
15,
15,
15,
15,
15,
15,
15,
15
135
15
B
11,
12,
13,
14,
15,
16,
17,
18,
19,
135
15
C
3,
6,
9,
12,
15,
18,
21,
24,
27
135
15
All the three series A, B and C, have the same size (n = 9) and same mean viz., 15. Thus, if we are
given that the mean of a series of 9 observations is 15, we cannot determine if we are talking of the series
A, B or C. In fact, any series of 9 items with total 135 will give mean 15. Thus, we may have a large
number of series with entirely different structures and compositions but having the same mean.
From the above illustration it is obvious that the measures of central tendency are inadequate to
describe the distribution completely. In the words of George Simpson and Fritz Kafka :
“An average does not tell the full story. It is hardly fully representative of a mass unless we know the
manner in which the individual items scatter around it. A further description of the series is necessary if we
are to gauge how representative the average is.”
Thus the measures of central tendency must be supported and supplemented by some other measures.
One such measure is Dispersion.
Literal meaning of dispersion is “Scatteredness.” We study dispersion to have an idea of the
homogeneity (compactness) or heterogeneity (scatter) of the distribution. In the above illustration, we say
that the series A is stationary, i.e., it is constant and shows no variability. Series B is slightly dispersed and
series C is relatively more dispersed. We say that series B is more homogeneous (or uniform) as compared
with series C or the series C is more heterogeneous than series B.
We give below some definitions of dispersion as given by different statisticians from time to time.
WHAT THEY SAY ABOUT DISPERSION — SOME DEFINITIONS
“Dispersion is the measure of the variation of the items.”—A.L. Bowley
“Dispersion is a measure of the extent to which the individual items vary.”—L.R. Connor
“Dispersion or spread is the degree of the scatter or variation of the variables about a central
value.” —B.C. Brooks and W.F.L . Dick
“The degree to which numerical data tend to spread about an average value is called the
variation or dispersion of the data.”—Spiegel
“The term dispersion is used to indicate the facts that within a given group, the items differ
from one another in size or in other words, there is lack of uniformity in their sizes.”—W.I. King
6·2
BUSINESS STATISTICS
6·1·1. Objectives or Significance of the Measures of Dispersion. The main objectives of studying
dispersion may be summarised as follows :
1. To find out the reliability of an average. The measures of variation enable us to find out if the
average is representative of the data. As stated earlier, dispersion gives us an idea about the spread of the
observations about an average value. If the dispersion is small, it means that the given data values are closer
to the central value (average) and hence the average may be regarded as reliable in the sense that it provides
a fairly good estimate of the corresponding population average. If the dispersion is large, then the data
values are more deviated from the central value, thereby implying that the average is not representative of
the data and hence not quite reliable.
2. To control the variation of the data from the central value. The measures of variation help us to
determine the causes and the nature of variation, so as to control the variation itself.
It helps to measure the extent of variation from the standard quality of various works carried in
industries. For example, we use 3-sigma (3-σ) control limits to determine if a manufacturing process is in
control or not. This helps us to identify the causes of variation in the manufactured product and accordingly
take corrective and remedial measures. [For detailed discussion, see Chapter 21, Statistical Quality
Control.] The Government can also take suitable policy decisions to remove the inequalities in the
distribution of income and wealth, after careful study of the dispersion of the income and wealth.
3. To compare two or more sets of data regarding their variability. The relative measures of dispersion
may be used to compare two or more distributions, even if they are measured in different units, as regards
their variability or uniformity. For detailed discussion, see § 6·12 Coefficient of Variation.
4. To obtain other statistical measures for further analysis of data. The measures of variation are used
for computing other statistical measures which are used extensively in Correlation Analysis (Chapter 8),
Regression Analysis (Chapter 9), Theory of Estimation and Testing of Hypothesis (Chapter 16), Statistical
Quality Control (Chapter 21) and so on.
6·2. CHARACTERISTICS FOR AN IDEAL MEASURE OF DISPERSION
The desideratta for an ideal measure of dispersion are the same as those for an ideal measure of central
tendency, viz. :
(i) It should be rigidly defined.
(ii) It should be easy to calculate and easy to understand.
(iii) It should be based on all the observations.
(iv) It should be amenable to further mathematical treatment.
(v) It should be affected as little as possible by fluctuations of sampling.
(vi) It should not be affected much by extreme observations.
All these properties have been explained in Chapter 5 on Measures of Central Tendency.
6·3. ABSOLUTE AND RELATIVE MEASURES OF DISPERSION
The measures of dispersion which are expressed in terms of the original units of a series are termed as
Absolute Measures. Such measures are not suitable for comparing the variability of the two distributions
which are expressed in different units of measurement. On the other hand, Relative Measures of dispersion
are obtained as ratios or percentages and are thus pure numbers independent of the units of measurement.
For comparing the variability of the two distributions (even if they are measured in the same units), we
compute the relative measures of dispersion instead of the absolute measures of dispersion.
6·4. MEASURES OF DISPERSION
The various measures of dispersion are :
(i) Range.
(ii) Quartile deviation or Semi-Interquartile range.
(iii) Mean deviation.
MEASURES OF DISPERSION
6·3
(iv) Standard deviation.
(v) Lorenz curve.
The first two measures viz., Range and Quartile deviation are termed as positional measures since they
depend upon the values of the variable at particular position of the distribution. The last measure viz.,
Lorenz curve is a graphic method of studying variability. In the following sections we shall discuss these
measures in detail one by one.
6·5. RANGE
The range is the simplest of all the measures of dispersion. It is defined as the difference between the
two extreme observations of the distribution. In other words, range is the difference between the greatest
(maximum) and the smallest (minimum) observation of the distribution. Thus
Range = Xmax – Xmin
(6·1)
where Xmax = L, is the greatest observation and Xmin = S, is the smallest observation of the variable values.
In case of the grouped frequency distribution (for discrete values) or the continuous frequency
distribution, range is defined as the difference between the upper limit of the highest class and the lower
limit of the smallest class.
Remarks 1. In case of a frequency distribution, the frequencies of the various values of the variable (or
classes) are immaterial since range depends only on the two extreme observations.
2. Absolute and Relative Measures of Range. Range as defined in (6·1) is an absolute measure of
dispersion and depends upon the units of measurement. Thus, if we want to compare the variability of two
or more distributions with the same units of measurement, we may use (6·1). However, to compare the
variability of the distributions given in different units of measurement, we cannot use (6·1) but we need a
relative measure which is independent of the units of measurement. This relative measure, called the
coefficient of range, is defined as follows :
X – Xmin
Coefficient of Range = max
…(6·2)
Xmax + Xmin
where the symbols have already been explained. In other words, coefficient of range is the ratio of the
difference between two extreme observations (the biggest and the smallest) of the distribution to their sum.
It is a common practice to use coefficient of range even for the comparison of variability of
distributions given in the same units of measurement.
3. Effect of Change of Origin and Scale on Range.
Let the given variable X be transformed to the new variable U by changing the origin and scale in X as
follows :
U = AX + B
…(6.3)
⇒
Umax =A . Xmax + B
and
Umin = A . Xmin + B
∴
Range (U) = Umax – Umin = A (Xmax – Xmin)
⇒
Range (U) = A . Range (X)
…(6.3a)
Hence, range is independent of change of origin but not of scale.
4. When is dispersion (variation) zero ?
A crude measure of dispersion is :
Range = Largest sample observation – Smallest sample observation.
Range = 0, if Largest sample observation = Smallest sample observation
This is possible only if the variable takes a constant value i.e., if all the observations in the sample have
the same value.
For example, if a variable takes 5 values 8, 8, 8, 8, 8, then :
Largest value (L) = 8 and Smallest value (S) = 8
∴
Range = L – S = 8 – 8 = 0.
6·4
BUSINESS STATISTICS
6·5·1. Merits and Demerits of Range. Range is the simplest though crude measure of dispersion. It is
rigidly defined, readily comprehensible and is perhaps the easiest to compute, requiring very little
calculations. However, it does not satisfy the properties (iii) to (vi) for an ideal measure of dispersion. We
give below its limitations and drawbacks.
(i) Range is not based on the entire set of data. It is based only on two extreme observations, which
themselves are subject to change fluctuations. As such, range cannot be regarded as a reliable measure of
variability.
(ii) Range is very much affected by fluctuations of sampling. Its value varies very widely from sample
to sample.
(iii) If the smallest and the largest observations of a distribution are unaltered and all other values are
replaced by a set of observations within these values i.e., X max and X min, the range of the distribution
remains same. Moreover if any item is added or deleted on either side of the extreme value, the value of the
range is changed considerably, though its effect is not so pronounced if we use the coefficient of range.
Thus range does not take into account the composition of the series or the distribution of the observations
within the extreme values. Consequently, it is fairly unreliable as a measure of dispersion of the values
within the distribution.
(iv) Range cannot be used if we are dealing with open end classes.
(v) Range is not suitable for mathematical treatment.
(vi) Another shortcoming of the range, though less important is that it is very sensitive to the size of the
sample. As the sample size increases, the range tends to increase though not proportionately.
(vii) In the words of W.I. King “Range is too indefinite to be used as a practical measure of
dispersion.”
6·5·2. Uses. (1) In spite of the above limitations and shortcomings range, as a measure of dispersion,
has its applications in a number of fields where the data have small variations like the stock market
fluctuations, the variations in money rates and rate of exchange.
(2) Range is used in industry for the statistical quality control of the manufactured product by the
construction of R-chart i.e., the control chart for range.
(3) Range is by far the most widely used measure of variability in our day-to-day life. For example, the
answer to problems like, ‘daily sales in a departmental store’; ‘monthly wages of workers in a factory’ or
‘the expected return of fruits from an orchard’, is usually provided by the probable limits - in the form of a
range.
(4) Range is also used as a very convenient measure by meteorological department for weather
forecasts since the public is primarily interested to know the limits within which the temperature is likely to
vary on a particular day.
Example 6·1. Calculate the range and the coefficient of range of A’s monthly earnings for a year.
Month
Monthly earnings
Month
Monthly earnings
Month
Monthly earnings
(In ’00 Rs.)
(In ’00 Rs.)
(In ’00 Rs.)
1
139
5
157
9
162
2
150
6
158
10
162
3
151
7
160
11
173
4
151
8
161
12
175
Solution.
Largest earning (L) = Rs. 17,500 ; Smallest earnings (S) = Rs. 13,900
∴
Range = L – S = 17,500 – 13,900 = Rs. 3,600.
L – S 17‚500 – 13‚900 36
=
=
= 0·115
Coefficient of range =
L + S 17‚500 + 13‚900 314
Example 6·2(a). The following table gives the age distribution of a group of 50 individuals.
Age (in years)
:
16 – 20
21 – 25
26 – 30
31 – 36
No. of persons
:
10
15
17
8
Calculate range and the coefficient of range.
6·5
MEASURES OF DISPERSION
(b). If the variables x and y are related by 3x – 2y + 5 = 0 and the range of x is 8, find the range of y .
[I.C.W.A. (Foundation), Dec. 2005]
Solution. (a) Since age is a continuous variable we should first convert the given classes into
continuous classes. The first class will then become 15·5—20·5 and the last class will become 30·5—35·5.
Largest value = 35·5 ; Smallest value = 15·5
∴
Range = 35·5 – 15·5 = 20 years
35·5 – 15·5
20
Coefficient of range = 35·5 + 15·5 = 51 = 0·39.
(b) We are given :
and
Range (x) = 8
3x – 2y + 5 = 0
…(1)
1
2
⇒
3
2
y = (3x + 5) = x +
5
2
Hence, using (6.3a), we get
3
3
Range (y) = 2 Range (x) = 2 × 8 = 12
[From (1)]
6·6. QUARTILE DEVIATION OR SEMI INTER-QUARTILE RANGE
It is a measure of dispersion based on the upper quartile Q3 and the lower quartile Q 1 .
Inter-quartile Range = Q 3 – Q1
…(6·4)
Quartile deviation is obtained from inter-quartile range on dividing by 2 and hence is also known as
semi inter-quartile range. Thus
Q – Q1
Quartile Deviation (Q.D.) = 3
…(6·5)
2
Q.D. as defined in (6·5) is only an absolute measure of dispersion. For comparative studies of
variability of two distributions we need a relative measure which is known as Coefficient of Quartile
Deviation and is given by :
(Q – Q1)/2 Q3 – Q1
Coefficient of Q.D. = 3
=
…(6·6)
(Q 3 + Q1)/2 Q3 + Q1
Remarks 1. The quartile deviation gives the average amount by which the two quartiles differ from
median. For a symmetrical distribution we have (c.f. Chapter 7).
Q + Q3
Q3 – Md = Md – Q 1
⇒
Md = 1
…(*)
2
i.e., median lies half way on the scale from Q1 to Q3. Thus for a symmetrical distribution we have :
Q – Q1
Q + Q1
Q.D. + Q1 = 3
+ Q1 = 3
= Md
[From (*)]
2
2
Q – Q1
Q + Q1
and
Q3 – Q.D. = Q3 – 3
= 3
= Md
[From (*)]
2
2
In other words, for a symmetrical distribution we have :
Q1 = Md – Q.D. and Q 3 = Md + Q.D.
…(**)
Since in a distribution, 25% of the observations lie below Q1 and 25% observations lie above Q3, 50%
of the observations lie between Q 1 and Q 3 . Therefore, using (**) we conclude that for a symmetrical
distribution Md ± Q.D. covers exactly 50% of the observations.
2. Rigorously speaking quartile deviation is only a positional average and does not exhibit any scatter
around an average. As such some statisticians prefer to call it a measure of partition rather than a measure
of dispersion.
6·6·1. Merits and Demerits of Quartile Deviation. Merits. Quartile deviation is quite easy to
understand and calculate. It has a number of obvious advantages over range as a measure of dispersion. For
example :
(a) As against range which was based on two observations only, Q.D. makes use of 50% of the data
and as such is obviously a better measure than range.
6·6
BUSINESS STATISTICS
(b) Since Q.D. ignores 25% of the data from the beginning of the distribution and another 25% of the
data from the top end, it is not affected at all by extreme observations.
(c) Q.D. can be computed from the frequency distribution with open end classes. In fact, Q.D. is the
only measure of dispersion which can be obtained while dealing with a distribution having open end
classes.
Demerits. (i) Q.D. is not based on all the observations since it ignores 25% of the data at the lower end
and 25% of the data at the upper end of the distribution. Hence, it cannot be regarded as a reliable measure
of variability.
(ii) Q.D. is affected considerably by fluctuations of sampling.
(iii) Q.D. is not suitable for further mathematical treatment.
Thus quartile deviation is not a reliable measure of variability, particularly for distributions in which
the variation is considerable.
6·7. PERCENTILE RANGE
This is a measure of dispersion based on the difference between certain percentiles. If P i is the ith
percentile and Pj is the jth percentile then the so-called i-j percentile range is given by :
i-j Percentile Range = Pj – Pi, (i < j)
…(6·7)
Thus i -j Semi-percentile Range is given by :
(Pj – Pi ) / 2, (i < j)
…(6·7a)
th
th
The commonly used percentile range is the one which corresponds to the 10 and 90 percentiles.
Thus taking i = 10 and j = 90 in (6·7), we get
10-90 Percentile Range = P90 – P10
…(6·8)
and
10-90 Semi-percentile Range = (P 90 – P10)/2
…(6·8a)
The above measures are absolute measures only. The relative measure of variability based on
percentiles is given by :
(P90 – P10)/2 P90 – P10
Coefficient of 10-90 percentile =
=
…(6·9)
(P90 + P10)/2 P90 + P10
Theoretically, 10-90 percentile range should serve as a better measure of dispersion than Q.D. since it
is based on 80% of the data. However, in practice it is not commonly used.
Example 6·3. Find :
(i) Inter-quartile Range;
(ii) Quartile Deviation;
(iii) Coefficient of Quartile Deviation,
for the following distribution :
Class Interval :
0—15
15—30
30—45
45—60
60—75
75—90
90—105
f
:
8
26
30
45
20
17
4
Solution.
COMPUTATION OF QUARTILES
Here N/4 = 37·5. The c.f. just greater than 37·5 is 64.
Hence, Q1 lies in the corresponding class 30—45.
h N
15
∴ Q1 = l +
– C = 30 + 30 37·5—34
f 4
(
)
(
)
= 30 + 3·5
2 = 30 + 1·75 = 31·75
3N/4 = 112·5. The c.f. just greater than 112·5 is 129.
Hence, Q3 lies in the corresponding class 60—75.
h 3N
15
∴ Q3 = l +
– C = 60 + 20
112·5 – 109
f 4
(
= 60 +
3 × 3·5
4 =
)
(
60 + 2·625 = 62·625
)
Class Interval
f
(Less than) c.f.
——————————————————————————
————————————————
—————————————————————————
0—15
15—30
30—45
45—60
60—75
75—90
90—105
8
26
30
45
20
17
4
8
34
64
109
129
146
150
——————————————————————————————————————————————————————————————————
Total
N = 150
6·7
MEASURES OF DISPERSION
(i) Inter-quartile Range = Q3 – Q1 = 62·625 – 31·750 = 30·875
Q – Q1 30·875
(ii) Quartile Deviation = 3
= 2 = 15·44
[From Pat (i)]
2
(iii) Using the results in Part (i), we get :
Q – Q1 62·63 – 31·75 30·88
Coefficient of Q.D. = 3
=
=
= 0·33.
Q3 + Q1 62·63 + 31·75 94·38
Example 6·4(a). Evaluate an appropriate measure of dispersion for the following data :
Income (in Rs.)
: Less than 50 50—70
70—90 90—110 110—130 130—150 Above 150
No. of persons
:
54
100
140
300
230
125
51
(b) Comment on the following :
If the coefficient of quartile deviation (Q.D) is 0.6 and Q.D. = 15, then Q 1 = 10 and Q3 = 40.
[Delhi Univ. B.Com (Hons.), 2009]
Solution. (a) Since we are given the classes with open end intervals, the only measure of dispersion
that we can compute is the quartile deviation.
COMPUTATION OF QUARTILE DEVIATION
N 1000
4 = 4
3N
4 =
Income (in Rs.) No. of persons (f)
Here N = 1000,
= 250,
750
54
Since c.f just greater than 250 is 294, Q1 (first Less than 50
50 —70
100
quartile) lies in the corresponding class 70—90.
70 —90
140
Similarly, since c.f. just greater than 750 is 824, the
corresponding class 110—130 contains Q3. Hence,
90 —110
300
110—130
230
20
96
Q1 = 70 + 140 (250 – 154) = 70 +
130—150
125
7
Above 150
51
= 70 + 13·714 = 83·714
20
Q3 = 110 + 230 (750 – 594) = 110 +
∴
Quartile deviation (Q.D.) =
2 × 156
23 =
Q.D. =
54
154
294
594
824
949
1000
110 + 13·565 = 123·565
Q3 – Q1 123·565 – 83·714 39·851
=
= 2 = 19·925.
2
2
(Q 3 – Q1)
= 15 (Given)
2
Q – Q1
Coefficient of Q.D. = 3
= 0.6
Q3 + Q1
Adding (1) and (2), we get 2Q3 = 80
Subtracting (1) from (2), we get 2Q1 = 20
Hence, the given statement is true.
(b)
Less than c.f.
————————————————————————
—————————————————————————
——————————————————————
⇒
Q3 – Q1 = 30
⇒
Q3 + Q1 =
⇒
⇒
Q3 = 40
Q1 = 10
30 30 × 10
=
= 50
0.6
6
…(1)
…(2)
EXERCISE 6·1
1. (a) “Frequency distributions may either differ in the numerical size of their averages though not necessarily in
their formations or they may have the same values of their averages yet differ in their respective formations.”
Explain and illustrate how the measures of dispersion afford a supplement to the information about the frequency
distributions given by the averages.
(b) Discuss the validity of the statement : “An average, when published, should be accompanied by a measure of
dispersion, for significant interpretation.”
2(a) Find the range and the coefficient of range for the following observations.
65, 70, 82, 59, 81, 76, 57, 60, 55 and 50.
[C.A. PEE-1, Nov. 2003]
Ans. 32 ; 0.2424
(b) From the monthly income of 10 families given below, calculate
(a) the median,
(b) the geometric mean,
(c) the coefficient of range.
6·8
BUSINESS STATISTICS
S. No.
:
1
2
3
4
5
6
7
8
9
10
Income in Rs. :
145
367
268
73
185
619
280
115
870
315
Ans. (a) Md = Rs. 274
(b) G = Rs. 252·4,
(c) Coefficient of Range = 0·84.
3. The index numbers of prices of cotton shares (I1) and coal shares (I2) in a given year are as under—
Month :
Jan. Feb.
March
April
May June
July August Sept. Oct. Nov. Dec.
173
164
172
183
184
185
211
217
232
240
I1
: 188 178
130
129
129
120
127
127
130
137
140
142
I2
: 131 130
Calculate range for each share. Hence, discuss which share do you consider more variable in price.
Ans. Range (I1) = 76, Coefficient of Range (I1) = 0·19 ; Range (I2) = 22, Coefficient of Range (I2) = 0·084.
Cotton shares are more variable in prices.
4. Find the value of third quartile if the values of first quartile and quartile deviation are 104 and 18 respectively.
[Delhi Univ. B.Com. (Pass), 2002]
Ans. Q3 = 140.
5. Age distribution of 200 employees of a firm is given below : Construct a ‘less than ogive curve, and hence or
Q 3 – Q1
otherwise calculate semi-interquartile range
of the distribution :
2
Age in years (less than)
:
25
30
35
40
45
50
55
No. of employees
:
10
25
75
130
170
189
200
Q 3 – Q1
Ans. Q1 = 33·5 years,
Q3 = 43 years,
= 4·75 years
2
6. Find the mode, median, lower quartile (Q1) and upper quartile (Q3) and Coeff. of Q.D. from the following data :
Wages
:
0—10
10—20
20—30
30—40
40—50
No. of workers
:
22
38
46
35
20
[Maharishi Dayanand Univ. B.Com., 1997]
Ans. Mode = 24·21 : Median = 24·46,
Q1 = 14·803,
Q3 = 24·21 ;
Coeff. of Q.D. = 0·396.
7. Compute the Coefficient of Quartile Deviation of the following data :
Size
Frequency
Size
Frequency
12
24—28
6
4—8
10
28—32
10
8—12
6
32—36
18
12—16
2
36—40
30
16—20
15
20—24
Ans. Q1 = 14·5,
Q3 = 24·92,
Coefficint of Q.D. = 0·2643.
8. Find (i) Inter-quartile range, (ii) Semi-inter-quartile range, and (iii) Coefficient of quartile deviation,
from the following frequency distribution :
Marks
:
10—20 20—30 30—40 40—50 50—60 60—70 70—80 80—90
No. of students :
60
45
120
25
90
80
120
60
[C.A. (Foundation), Dec. 1993]
Ans. (i) 38·75,
(ii) 19·375,
(iii) 0·3647.
9. From the following data,
(i) Calculate the ‘percentage’ of workers getting wages : (a) more than Rs. 44 ; (b) between Rs. 22 and Rs. 58.
(ii) Find the quartile deviation.
Wages (Rs.)
:
0—10
10—20 20—30 30—40 40—50 50—60 60—70 70—80
No. of workers
:
20
45
85
160
70
55
35
30
Ans. (i) (a) 32·4%, (b) 68·4%; (ii) Q1 = 27·06, Q3 = 49·29, Q.D. = 11·115.
10. Calculate the appropriate measure of dispersion from the following data :
Wages in Rs. per week :
Less than 35
35—37
38—40
41—43
Over 43
No. of wage earners :
14
62
99
18
7
Ans. Coefficient of Q.D. = 0·046.
11. Find out middle 50%, middle 80% and coefficient of Q.D. from the following table :
Size of items
:
2
4
6
8
10
12
Frequency
:
3
5
10
12
6
4
Ans. Quartile range = 4 ; Percentile range = 8, Coefficient of Q.D. = 0·25.
6·9
MEASURES OF DISPERSION
6·8. MEAN DEVIATION OR AVERAGE DEVIATION
As already pointed out, the two measures of dispersion discussed so far viz., range and quartile
deviation are not based on all the observations and also they do not exhibit any scatter of the observations
from an average and thus completely ignore the composition of the series. Mean Deviation or the Average
Deviation overcomes both these drawbacks. As the name suggests, this measure of dispersion is obtained
on taking the average (arithmetic mean) of the deviations of the given values from a measure of central
tendency. According to Clark and Schkade :
“Average deviation is the average amount of scatter of the items in a distribution from either the mean
or the median, ignoring the signs of the deviations. The average that is taken of the scatter is an arithmetic
mean, which accounts for the fact that this measure is often called the mean deviation.”
6·8·1. Computation of Mean Deviation. If X 1 , X 2 ,…,Xn are n given observations then the mean
deviation (M.D.) about an average A, say, is given by :
M.D. (about an average A) = 1n ∑ | X – A | = n1 ∑ | d |
…(6·10)
where | d | = | X – A | read as mod (X – A), is the modulus value or absolute value of the deviation (after
ignoring the negative sign) d = X – A and ∑ | d | is the sum of these absolute deviations and A is any one of
the averages Mean (M), Median (Md) and Mode (Mo).
Steps for Computation of Mean Deviation
1. Calculate the average A of the distribution by the usual methods.
2. Take the deviation d = X – A of each observation from the average A.
3. Ignore the negative signs of the deviations, taking all the deviations to be positive to obtain the
absolute deviations, | d | = | X – A |.
4. Obtain the sum of the absolute deviations obtained in step 3.
5. Divide the total obtained in step 4 by n, the number of observations.
The result gives the value of the mean deviation about the average A.
In the case of frequency distribution or grouped or continuous frequency distribution, mean deviation
about an average A is given by :
1
1
…(6·11)
M.D. (about the average A) = ∑ f | X – A | = ∑ f | d |
N
N
where X is the value of the variable or it is the mid-value of the class interval (in the case of grouped or
continuous frequency distribution), f is the corresponding frequency, N = ∑f , is the total frequency and
| X – A | is the absolute value of the deviation d = (X – A) of the given values of X from the average A
(Mean, Median or Mode).
Steps for Computation of Mean Deviation for Frequency Distribution
Steps 1, 2 and 3 are same as given above.
4. Multiply the absolute deviations | d | = | X – A | by the corresponding frequency f to get f | d |.
5. Take the total of products in step 4 to obtain ∑ f | d |.
6. Divide the total in step 5 by N, the total frequency.
The resulting value is the mean deviation about the average A.
Remarks 1. Usually, we obtain the mean deviation (M.D.) about any one of the three averages
mean (M), median (Md) or mode (Mo). Thus
M.D. (about mean)
= 1N ∑ f | X – M |
1
M.D. (about median) = N
∑ f | X – Md |
M.D. (about mode)
1
=N
∑
f | X – Mo |
}
…(6·11a)
2. The sum of the absolute deviations (after ignoring the signs) of a given set of observations is
minimum when taken about median. Hence mean deviation is minimum when it is calculated from median.
6·10
BUSINESS STATISTICS
In other words, mean deviation calculated about median will be less than mean deviation about mean or
mode.
3. As already pointed out in Remark 1, usually, we compute the mean deviation about any one of the
three averages mean, median or mode. But since mode is generally ill-defined, in practice M.D. is
computed about mean or median. Further, as a choice between mean and median, theoretically, median
should be preferred since M.D. is minimum when calculated about median (c.f. Remark 2). But because of
wide applications of mean in Statistics as a measure of central tendency, in practice mean deviation is
generally computed from mean.
4. For a symmetrical distribution the range Mean ± M.D. (about mean) or Md ± M.D. (about median),
[·. · M = M d for a symmetrical distribution] covers 57·5% of the observations of the distribution. If the
distribution is moderately (skewed), the range will cover approximately 57·5% of the observations.
5. Effect of Change of Origin and Scale on Mean Deviation About Mean.
Let X1 , X2 , …, X n be n observations on the variable X. Let the variable X be transformed to the new
variable Y by changing the origin and scale in X, by the transformation :
—
—
—
—
Y = AX + B
⇒
Y = AX + B
⇒
Y – Y = A (X – X )
…(6.12)
By definition,
—
—
1 n
1 n
M.D. (Y) about mean = ∑ | (Yi – Y ) | = ∑ | A (Xi – X ) |
[From (6.12)]
n i=1
n i=1
=|A|.
—
1 n
∑ | X – X | = | A | . [M.D. (X) about mean]
n i=1 i
Hence, we have the following result.
Y = AX + B
⇒
M.D. (Y) about mean = | A | × [M.D. (X) about mean]
…(6.12a)
6·8·2. Short-cut Method of Computing Mean Deviation. For computing the mean deviation, first of
all we have to calculate the average about which we want the mean deviation, by the methods discussed in
the previous chapter. In case the average is a whole number, the method of computing mean deviation by
formulae (6·10), (6·11) or (6·12) is quite convenient. But if the average comes out in fractions, this method
becomes fairly tedious and time-consuming and requires lot of algebraic calculations. In such a case we
compute the mean deviation by taking the deviations from an arbitrary point ‘a’, near the average value and
applying some corrections as given below :
[
1
M.D. (about Mean) = N
∑ f | X – a | + (M – a) ( ∑ fB – ∑ fA )
]
…(6·13)
where
and
M is the mean,
a is the arbitrary constant near the mean,
∑ fB is the sum of all the class frequencies before and including the mean value,
∑ fA is the sum of all the class frequencies after the mean value.
Similarly, using the short cut method,
1
M.D. (about median) = N
[ ∑ f | X – a | + (Md – a) (∑ f ′ – ∑ f ′) ]
B
A
…(6·13a)
where now, Md is the median, a is some arbitrary constant near the median, ∑ fB′ is the sum of the class
frequencies before and including the median value, and ∑ fA′ is the sum of the class frequencies after the
median value.
Remarks 1. Obviously,
∑fA + ∑ fB = N
…(6·14)
⇒
∑ fB = N – ∑ fA
or
∑ fA = N – ∑ fB
Similarly
∑ fB′ = N – ∑ fA′
…(6·14a)
2. The above formulae (6·13) and (6·13a) are true provided all the values of the variable which are
above the average (M or Md) are also above ‘a’ and those which are below the average are also below ‘a’.
MEASURES OF DISPERSION
6·11
The arbitrary constant ‘a’ should be taken some arbitrary integral value near the average value, i.e., it
should be a value in the average class. The short cut method will not yield correct result if ‘a’ is taken
outside the average class.
6·8·3. Merits and Demerits of Mean Deviation
Merits : (i) Mean deviation is rigidly defined and is easy to understand and calculate.
(ii) Mean deviation is based on all the observations and is thus definitely a better measure of dispersion
than the range and quartile deviation.
(iii) The averaging of the absolute deviations from an average irons out the irregularities in the
distribution and thus mean deviation provides an accurate and true measure of dispersion.
(iv) As compared with standard deviation (discussed in next article § 6·9), it is less affected by extreme
observations.
(v) Since mean deviation is based on the deviations about an average, it provides a better measure for
comparison about the formation of different distributions.
Demerits. (i) The strongest objection against mean deviation is that while computing its value we take
the absolute value of the deviations about an average and ignore the signs of the deviations.
(ii) The step of ignoring the signs of the deviations is mathematically unsound and illogical. It creates
artificiality and renders mean deviation useless for further mathematical treatment. This drawback
necessitates the requirement of another measure of variability which, in addition to being based on
all the observations is also amenable to further algebraic manipulations.
(iii) It is not a satisfactory measure when taken about mode or while dealing with a fairly skewed
distribution. As already pointed out, theoretically mean deviation gives the best result when it is
calculated about median. But median is not a satisfactory measure when the distribution has great
variations.
(iv) It is rarely used in sociological studies.
(v) It cannot be computed for distributions with open end classes.
(vi) Mean deviation tends to increase with the size of the sample though not proportionately and not so
rapidly as range.
6·8·4. Uses. In spite of its mathematical drawbacks, mean deviation has found favour with economists
and business statisticians because of its simplicity, accuracy and the fact that standard deviations (discussed
in § 6·9) gives greater weightage to the deviations of extreme observations. Mean deviation is frequently
useful in computing the distribution of personal wealth in a community or a nation since for this, extremely
rich as well as extremely poor people should be taken into consideration. Regarding the practical utility of
mean deviation as a measure of variability, it may be worthwhile to quote that in the studies relating to
forecasting business cycles, the National Bureau of Economic Research has found that the mean deviation
is most practical measure of dispersion to use for this purpose.
6·8·5. Relative Measures of Mean Deviation. The measures of mean deviation as defined in (6·10),
(6·11) and (6·12) are absolute measures depending on the units of measurement. The relative measure of
dispersion, called the coefficient of mean deviation is given by :
Mean Deviation
…(6·15)
Coefficient of M.D. =
Average about which it is calculated
M.D.
∴
Coefficient of M.D. about mean =
…(6·15a)
Mean
M.D.
and
Coefficient of M.D. about median =
…(6·15b)
Median
The coefficients of mean deviation defined in (6·15), (6·15a) and (6·15b) are pure numbers independent
of the units of measurement and are useful for comparing the variability of different distributions.
Example 6·5. Calculate the mean deviation from mean for the following data.
Class Interval
:
2—4
4—6
6—8
8—10
Frequency
:
3
4
2
1
6·12
BUSINESS STATISTICS
Solution.
Class
Mid-Value
(X)
2—4
4—6
6—8
8—10
3
5
7
9
COMPUTATION OF MEAN AND M.D. FROM MEAN
—
—
Frequency
d=X–5
fd
|X– X|
f|X–X|
(f)
= | X – 5·2 |
3
–2
–6
2·2
6·6
4
0
0
0·2
0·8
2
2
4
1·8
3·6
1
4
4
3·8
3·8
—
∑ f = 10
∑ fd = 2
∑ f | X – X | = 14·8
—
X =A+
6
0
4
4
∑ f | d | = 14
∑ fd
2
= 5 + = 5·2
N
10
—
1
f | d |*
M.D. about mean = N ∑ f | X – X | =
14·8
10 =
1·48
*
Last column f | d | is not required for this method. It is needed for the short-cut method given below.
Aliter. Short-cut Method. We can use the deviations from arbitrary point a = 5, directly to compute
—
—
the M.D. from mean without computing the values | X – X |. This is particularly useful when X is in
fractions (decimals) in which case the usual formula is quite laborious. For this we need the last column
f | d |. [c.f. (6·13), § 6·8·2]. Using the formula (6·13), we get
1
[
—
M.D. about mean = N ∑ f | d | + ( X – a) (∑ fB – ∑ fA)
]
where
∑ fB = Sum of all the class frequencies before and including the mean class i.e., the class in which
mean lies.
Here mean is 5·2 and it lies in the class 4—6.
∴
∑ fB = 4 + 3 = 7
∑ fA = Sum of all the class frequencies after the mean class = 2 + 1 = 3
or
∑ fA = N – ∑ fB = 10 – 7 = 3.
∴
1
[
]
M.D. about mean = 10 14 + (5·2 – 5) × (7 – 3) =
14 + 0·8
10
=
14·8
10
= 1·48.
Remark. The value obtained by the short-cut method coincides with the value obtained by the direct
method and rightly so because the arbitrary point A = 5 is near the mean value 5·2 (c.f. Remark 2, § 6·8·2).
Example 6·6. (a) Find the Mean Deviation from the Mean for the following data :
Class Interval :
0—10
10—20
20—30
30—40
40—50
50—60
60—70
Frequency
:
8
12
10
8
3
2
7
[C.A. (Foundation), Nov. 1996]
(b) Also find the mean deviation about median.
(c) Compare the results obtained in (a) and (b).
Solution.
Class
Interval
0—10
10—20
20—30
30—40
40—50
50—60
60—70
Total
CALCULATIONS FOR M.D. ABOUT MEAN AND MEDIAN
—
—
Mid-value Frequency Less than
fX
| X – Md |
|X– X|
f|X–X|
(X)
(f)
c.f.
= | X – 22 |
= | X – 29 |
5
15
25
35
45
55
65
8
12
10
8
3
2
7
N = 50
8
20
30
38
41
43
50
40
180
250
280
135
110
455
∑fX
= 1,450
24
14
4
6
16
26
36
192
168
40
48
48
52
252
—
∑ f | X– X |
= 800
17
7
3
13
23
33
43
f | X – Md |
136
84
30
104
69
66
301
∑ f | X – Md |
= 790
6·13
MEASURES OF DISPERSION
1
1‚450
∑ fX = 50 = 29.
N
—
1
800
Mean Deviation about mean = ∑ f | X – X | =
= 16.
N
50
(b) (N/2) = (50/2)= 25. The c.f. just greate than 25 is 30. Hence, the corresponding class 20—30 is the
median class. Using the median formula, we get
h N
10
Md = l +
– C = 20 + 25 (25 – 20) = 20 + 2 = 22
f 2
1
790
∴
Mean Deviation about median = ∑ f | X – Md | = 50 = 15·8
N
(c) From (a) and (b), we observe that :
M.D. about median < M.D. about mean.
In fact, we have the following general result :
“Mean deviation is least when taken about median.”
Example 6·7. Calculate mean deviation from the median for the following data :
Marks less than :
80
70
60
50
40
30
20
10
No. of Students
:
100
90
80
60
32
20
13
5
Solution. First of all we shall convert the given cumulative frequency distribution table into ordinary
frequency distribution as given in the following table.
—
(a).
Mean ( X ) =
(
Marks
c.f.
0—10
10—20
20—30
30—40
40—50
50—60
60—70
70—80
5
13
20
32
60
80
90
100
)
COMPUTATION OF M.D. FROM MEDIAN
Frequency (f)
Mid-value of class (X)
| X – Md |
5
13 – 5 = 8
20 – 13 = 7
32 – 20 = 12
60 – 32 = 28
80 – 60 = 20
90 – 80 = 10
100 – 90 = 10
N = ∑ f = 100
5
15
25
35
45
55
65
75
41·43
31·43
21·43
11·43
1·43
8·57
18·57
28·57
f | X – Md |
207·15
251·44
150·00
137·16
40·04
171·14
185·70
285·70
∑ f | X – Md | = 1428·6
Here N/2 = 50. Since the c.f. just greater than 50 is 60, the corresponding class 40—50 is the median
class.
h N
10
∴
Md = l +
– C = 40 + 28 (50 – 32) = 40 + 6·43 = 46·43.
f 2
1
1428·6
∴
M.D. about Md = ∑ f | X – Md | =
= 14·286 ~– 14·29.
N
100
Aliter. Since median value comes out to be in fractions, we can do the above question conveniently by
the short-cut method i.e., by taking the deviations from any arbitrary point a = 45, (say), lying in the
median class.
(
Marks (X)
5
15
25
35
45
55
65
75
)
MEAN DEVIATION BY SHORT-CUT METHOD
f
d = X – 45
|d|
5
– 40
40
8
–30
30
7
–20
20
12
–10
10
28
0
0
20
10
10
10
20
20
10
30
30
∑ f = 100
f|d|
200
240
140
120
0
200
200
300
∑ f | d | = 1400
6·14
BUSINESS STATISTICS
Using formula (6·13a), we get
1
M.D. about Md =
∑ f | d | + (Md – a) (∑fB′ – ∑ fA′
N
where
∑ fB′ = Sum of the frequencies before and including the median value viz., 46·3.
= 5 + 8 + 7 + 12 + 28 = 60
∑ fA′ = N – ∑ fB′ = 100 – 60 = 40
[
∴
[
=[
M.D. about Md =
]
1400 + (46·43 – 45) (60 – 40)
100
1400 + 28·6
100
]
1428·6
= 100
][
1400 + 1·43 × 20
100
]
= 14·286 ~– 14·29.
Remark. As in the last question, the values of M.D. obtained by the direct method and the short-cut
method are same, since the arbitrary value ‘a’ is taken in the median class.
Example. 6.8. If 2xi + 3y i = 5; i = 1, 2, …, n and mean deviation of x1 , x2, …, xn about their mean is 12,
find the mean deviation of y1, y2, …, yn about their mean.
[I.C.W.A. (Foundation), Dec. 2006]
Solution. M.D. (X) about mean = 12 (Given).
(*)
2
5
Also
2xi + 3yi = 5
⇒
yi = – xi – ; i = 1, 2, …, n
…(**)
3
3
We know that if
Y = AX + B, then M.D. of (Y) about mean = | A | × [M.D. (X) about mean],
[From 6.12a]
where | A | is the modulus value of A.
∴
M.D. (Y) about mean = ⎪–
2
3⎪
× [M.D. (X) about mean ] = 32 × 12 = 8
[From (*)]
Example 6·9. Mean deviation may be calculated from the arithmetic mean or the median or the mode.
Which of these three measures is the minimum ?
Five towns A, B, C, D and E lie in that order along a road. The distances in kilometres of the towns as
measured from A are A —0, B—5, C—9, D—16, E—20, A new college is to be established at one of these 5
places, and the number of students who would join the college are A—39, B—911, C—46, D—193, and
E—716. The criterion for choosing the location is that the total distance travelled by the students, as
measured in student-kilometres should be a minimum. (Thus if the college is located at D, 46 students from
C will travel in all 46 × 7 = 322 student-kilometres). Where should the college be located ? Justify your
result.
Solution. Mean deviation calculated from the median is the minimum.
Let X denote the distance (in kms.) as
Town
Distance from
No. of
Less than
measured from the town A. Then the frequency
town A
students
c.f.
distribution of the distances covered by the
(X)
(f)
students in going to the college is as given in the
A
0
39
39
adjoining table.
B
5
911
950
We are given that the criterion for choosing the
C
9
46
996
location of the college is that the total distance
D
16
193
1189
travelled by the students, as measured in studentE
20
716
N = 1905
kilometres should be minimum. Since mean
————————————————————————————————————————————————————————————
——————————————————
deviation is minimum when calculated from median, the total distance travelled by the students, as
measured in student-kilometres will be minimum at the point X = Median, of the frequency distribution of
X.
Here (N/2) = (1905/2) = 952·5. The c.f. just greater than 952·5 is 996. Hence, the corresponding value
of X = 9, is the median. Since X = 9, corresponds to the town C, the college should be located at the town C.
6·15
MEASURES OF DISPERSION
EXERCISE 6·2
1. What do you mean by ‘mean deviation’. Discuss its relative merits over range and quartile deviation as a
measure of dispersion. Also point out its limitations.
2. Calculate mean deviation about A.M. from the following :
Value (x)
:
10
11
12
13
Frequency (f)
:
3
12
18
12
Ans. A.M. = 11·87 ; M.D. = 0·71.
3. Calculate the mean deviation about meadian of the series :
x
2·5
3·5
4·5
5·5
6·5
7·5
8·5
9·5
10·5
f
2
3
5
6
6
4
6
4
14
Ans. M.D. (about median) = 2·22.
4. Compute the quartile deviation and mean deviation from median for the following data.
Height in inches
58
59
60
61
62
No. of students
15
20
32
35
33
Height in inches
63
64
65
66
—
No. of students
22
22
10
8
—
Ans. Q.D. = 1·5 ; M.D. (about median) = 1·73.
5. With median as the base, calculate the mean deviation and compare the variability of the two series A and B.
Series A :
3484
4572
4124
3682
5624
4388
3680
4308
Series B :
487
508
620
382
408
266
186
218
Ans. Series A : Md = 4216 ;
Series B : Md = 395 ;
M.D. = 490·25 ;
M.D. = 121·38 ;
Coeff. of M.D. = 0·116;
Coeff. of M.D. = 0·307. Series B is more variable.
6. Compare the dispersion of the following series by using the co-efficient of mean deviation.
Age (years)
:
16
17
18
19
20
21
22
23
No. of boys
:
4
5
7
12
20
13
5
0
No. of girls
:
2
0
4
8
15
10
6
3
24
4
2
Total
70
50
Ans. Coeff. of M.D. about median (boys) = 0·0685 ; Coeff. of M.D. about median (girls) = 0·0630.
7. Calculate the mean deviation from the mean for the following data :
Marks
:
0—10
10—20
20—30
30—40
No. of Students
:
6
5
8
15
40—50
7
Ans. Mean = 33·4 ; M.D. about mean = 13·184.
50—60
6
60—70
3
[C.A. (Foundation), May 1999]
8. (a) Mean deviation may be calculated from the arithmetic mean or the median or the mode ? Which of these
three measures is the minimum ?
(b) Find out mean deviation and its coefficient from median from the following series :
Size of items
:
4
6
8
10
12
14
16
Frequency
:
2
1
3
6
4
3
1
Ans. 2·4 ; 0·24
9. Calculate the mean deviation about the mean for the following data :
x
:
5
15
25
35
f
:
8
12
10
8
45
3
55
65
2
7
[C.A. (Foundation), May 2001]
Also find the M.D. about median and comment on the results obtained in (a) and (b).
Ans. Mean = 29; M.D. about mean = 16. ;
Median = 22 ; M.D. about median = 15·8.
6·16
BUSINESS STATISTICS
10. Calculate mean deviation from median from the following data :
Class interval
(f)
Class interval
50—55
6
20—25
55—60
12
25—30
60—70
17
30—40
70—80
30
40—45
10
45—50
Also calculate coefficient of mean deviation.
Ans. 8·75 ; 0·206.
(f)
10
8
5
2
11. The following distribution gives the difference in age between husband and wife in a particular community :
Difference in years :
0—5
5—10
10—15 15—20 20—25 25—30 30—35 35—40
Frequency
:
449
705
507
281
109
52
16
4
Calculate mean deviation about median from these data. What light does it throw on the social conditions of a
community ?
Ans. M.D. about median = 5·24.
12. Find the median and mean deviation of the following data :
Size
:
0—10
10—20
20—30
30—40
40—50
50—60
60—70
Frequency
:
7
12
18
25
16
14
8
Ans. Median = 35·2 ; M.D. = 13·148.
[Mysore Univ. B.Com., 1998]
13. Calculate the value of coefficient of mean deviation (from median) of the following data :
Marks
No. of Students
Marks
No. of Students
25
50—60
2
10—20
20
60—70
6
20—30
10
70—80
12
30—40
7
80—90
18
40—50
Ans. Median = 54·8 ; M.D. about median = 12·95; Coefficient of M.D. = 0·2363.
14. Compute the mean deviation from the median and from mean for the following distribution of the scores of 50
college students.
Score
:
140—150
150—160
160—170
170—180
180—190
190—200
Frequency :
4
6
10
10
9
3
Ans. 10·24 ; 10·56.
15. Calculate Mean Deviation from Median from the following data :
Wages in Rs. (Mid-value)
:
125
175
225
275
325
No. of persons
:
3
8
21
6
2
Ans. Median = 221·43 ; M.D. (Median) = 31·607.
6·9. STANDARD DEVIATION
Standard deviation, usually denoted by the letter σ (small sigma) of the Greek alphabet was first
suggested by Karl Pearson as a measure of dispersion in 1893. It is defined as the positive square root of the
arithmetic mean of the squares of the deviations of the given observations from their arithmetic mean. Thus
if X1 , X2 ,…, Xn is a set of n observations then its standard deviation is given by :
σ =
—
1
∑ (X – X )2
n
…(6·16)
1
X = ∑ X, is the arithmetic mean of the given values.
n
Steps for Computation of Standard Deviation
where
—
—
1. Compute the arithmetic mean X by the formula (6·16a).
—
2. Compute the deviation (X – X ) of each observation from arithmetic mean i.e., obtain
—
—
—
X1 – X , X2 – X , …, Xn – X .
…(6·16a)
6·17
MEASURES OF DISPERSION
—
—
—
3. Square each of the deviations obtained in step 2 i.e., compute (X1 – X )2, (X2 – X )2,…, (Xn – X )2.
4. Find the sum of the squared deviations in step 3 given by :
—
—
—
—
∑ (X – X )2 = (X 1 – X ) 2 + (X2 – X )2 + … + (Xn – X )2.
—
1
5. Divide this sum in step 4 by n to obtain ∑ (X – X )2.
n
6. Take the positive square root of the value obtained in step 5.
7. The resulting value gives the standard deviation of the distribution.
In case of frequency distribution, the standard deviation is given by :
σ =
—
1
∑ f(X – X )2
N
…(6·17)
where X is the value of the variable or the mid-value of the class (in case of grouped or continuous
frequency distributions) ; f is the corresponding frequency of the value X ; N = ∑ f, is the total frequency
and
—
1
X = ∑fX
… (6·17a)
N
is the arithmetic mean of the distribution.
Steps for Computation of Standard Deviation in case of Frequency Distribution
—
1. Compute X by the formula (6·17a) or the usual step deviation formula discussed in Chapter 5.
—
2. Compute deviations (X – X ) from the mean for each value of the variable X.
—
3. Obtain the squares of the deviations obtained in step 2 i.e., compute (X – X )2.
4. Multiply each of the squared deviations obtained in step 3 by the corresponding frequency to get
—
f (X – X )2.
—
5. Find the sum of the values obtained in step 4 to get ∑ f (X – X )2.
6. Divide the sum obtained in step 5 by N = ∑ f, the total frequency.
7. The positive square root of the value obtained in step 6 gives the standard deviation of the
distribution.
Remarks 1. It may be pointed out that although mean deviation could be calculated about any one of
the averages (M, Md or Mo), standard deviation is always computed about arithmetic mean.
2. To be more precise, the standard deviation of the variable X will be denoted by σx. This notation will
be useful when we have to deal with the standard deviation of two or more variables.
3. Standard deviation abbreviated as S.D. or s.d. is always taken as the positive square root in (6·16) or
(6·17).
—
—
—
4. The value of s.d. depends on the numerical value of the deviations (X1 –X ), (X2 – X ), …, (Xn – X ).
Thus the value of σ will be greater if the values of X are scattered widely away from the mean. Thus a small
value of σ will imply that the distribution is homogeneous and a large value of σ will imply that it is
heterogeneous. In particular s.d. is zero if each of the deviations is zero i.e., σ = 0 if and only if,
—
—
—
—
X1 – X = 0, X2 – X = 0, …, Xn – X = 0
⇒
X1 = X2 = X3 = … = Xn = X ,
which is the case if the variable assumes the constant value.
Thus, σ = 0 if and only if,
X1 = X2 = X3 = … = Xn = k (constant)
…(6·17b)
In other words, σ = 0 if and only if all the observations are equal.
As an illustration, let us suppose that the variable X takes the same constant values, say, 6, 6, 6, 6, 6.
Then
X
6
6
6
6
6
— 1
—
X = 5 ∑X = 6 + 6 + 56 + 6 + 6 = 30
5 =6
X– X =X–6
0
0
0
0
0
—————————————————————————————————————————————————————————————————-
∴
S.D. (σ) =
1
5
—
∑ (X – X )2 =
1
5
(0)
=0
6·18
BUSINESS STATISTICS
6·9·1. Mathematical Properties of Standard Deviation. Standard deviation possesses a number of
interesting and important mathematical properties which are given below.
1. Standard deviation is independent of change of origin but not of scale.
If d = X – A,
then
σx = σ d
X – A,
But
if d =
h > 0,
then
σx = h . σd.
h
2. Standard deviation is the minimum value of the root mean square deviation (§ 6·9·3)
3.
S.D. ≤ Range
i.e.,
σ ≤ Xmax – Xmin
[For Proof, see Remark to § 6·9·2.]
4. Standard deviation is suitable for further mathematical treatment. If we know the sizes, means and
standard deviations of two or more groups, then we can obtain the standard deviation of the group obtained
on combining all the groups. [For details see § 6·10]
5. The standard deviation of the first n natural numbers viz., 1, 2, 3, …, n is ⎯
√⎯⎯⎯⎯⎯⎯⎯
(n2 – 1)/12
[For Proof, see Example 6·23(a).]
6. The Empirical Rule. For a symmetrical bell shaped distribution, we have approximately the
following area properties.
(i) 68% of the observations lie in the range : Mean ± 1·σ.
(ii) 95% of the observations lie in the range : Mean ± 2.σ.
(iii) 99% of the observations lie in the range : Mean ± 3.σ..
7. The approximate relationship between quartile deviation (Q.D.), mean deviation (M.D.) and
standard deviation (σ) is :
2
4
Q.D. –~ 3 σ
and
M.D. ~– 5 σ
⇒
Q.D. : M.D. : S.D. : : 10 : 12 : 15
8. For any discrete distribution, standard deviation is not less than mean deviation about mean i.e.,
S.D. (σ) ≥ Mean Deviation about mean.
6·9·2. Merits and Demerits of Standard Deviation
Merits. Standard deviation is by far the most important and widely used measure of dispersion. It is
—
rigidly defined and based on all the observations. The squaring of the deviations (X —X ) removes the
drawback of ignoring the signs of deviations in computing the mean diviation. This step renders it suitable
for further mathematical treatment. The variance of the pooled (combined) series is given by formula (6·30)
in § 6·10.
Moreover, of all the measures of dispersion, standard deviation is affected least by fluctuations of
sampling.
Thus, we see that standard deviation satisfies almost all the properties laid down for an ideal measure
of dispersion except for the general nature of extracting the square root which is not readily comprehensible
for a non-mathematical person. It may also be pointed out that standard deviation gives greater weight to
extreme values and as such has not found favour with economists or businessmen who are more interested
in the results of the modal class. Taking into consideration the pros and cons and also the wide applications
of standard deviation in statistical theory, such as in skewness, kurtosis, correlation and regression analysis,
sampling theory and tests of significance, we may regard standard deviation as the best and the most
powerful measure of dispersion.
—
Remark. Since X – X ≤ R (Range), for all values of X viz., X1, X2 , …, Xn , we get
–
1
σ2 = ∑ f (X – X)2
N
–
–
–
1
1
=
f (X – X)2 + f2(X2 – X)2 + … + fn (Xn – X)2 ≤
f R2 + f2R2 + …+ fn R2
N 1 1
N 1
[
]
–
[
–
]
–
(·. ·X1 – X ≤ R , X2 – X ≤ R,…, Xn – X ≤ R)
6·19
MEASURES OF DISPERSION
1
1
· R2 (f1 + f2 + … + fn ) = R2 . N = R2
N
N
⇒
σ2 ≤ R2
⇒
σ≤R
⇒
s.d. ≤ Range.
6·9·3. Variance and Mean Square Deviation. Variance is the square of the standard deviation and is
denoted by σ2 . For a frequency distribution, variance is given by :
–
1
σ2 = ∑ f(X – X )2
…(6·18)
N
where the symbols have already been explained in (6·17).
The mean square deviation, usually denoted by s2 is defined as
1
s2 = ∑ f(X – A)2
…(6·19)
N
where A is any arbitrary number.
The square root of the mean square deviation is called root mean square deviation and given by :
∴
σ2 ≤
s =
1
∑ f (X – A)2
N
…(6·20)
Relation between σ 2 and s2 . We have
s2 ≥ σ 2 ⇒ s ≥ σ
…(6·21)
In other words, mean square deviation is not less than the variance or the root mean square deviation
is not less than the standard deviation.
—
The sign of equality will hold in (6·21) i.e., s 2 = σ2 if and only if X = A
…(6·22)
—
s2
Thus, will be least when X = A. Hence, mean square deviation or equivalently root mean square
deviation is least when deviations are taken from the arithmetic mean and variance (standard deviation) is
the minimum value of mean square deviation (root mean square deviation).
Important Remark
The variance σ2 is in squared units. For example, for the distribution of heights (in inches) of a group
of individuals, σ 2 is expressed in (inches)2 , a concept which is difficult to visualise and interpret. To
overcome this problem, we try to measure the variation in the sample data in the same units as those of the
original measurements by calculating the standard deviation (S.D.), which is defined as the positive square
root of variance. Hence, in practice, we use standard deviation, rather than variance as the basic unit of
variability.
6·9·4. Different Formulae for Calculating Variance. By definition, the variance of the random
variable X denoted by σ2 or more precisely by σx2, is given by
—
1
σx2 = ∑ f (X – X ) 2
…(·6·23)
N
where ∑ f = N, is the total frequency.
If X is not a whole number but comes out to be in fractions, the computation of σ x2 by the above
formula becomes very cumbersome and time-consuming. In order to overcome this difficulty we shall
develop different versions of the formula (6·23) which reduce the arithmetical calculations to a great extent
and are very useful for numerical computation of standard deviation.
Formula 1.
σx2 =
—
1
1
∑ f X2 – X 2 = ∑ f X2 –
N
N
(
1
∑fX
N
)
2
…(6·24)
Formula 2. If d = X – A , where A is arbitrary constant, then
σx2 = σd 2 =
1
∑ f d2 –
N
( N1 ∑ f d )
2
…(6·25)
6·20
BUSINESS STATISTICS
(6·24) is a much convenient form to use than the formula (6·23). But if the values of X and f are large, the
computation of f X, f X2 is quite time- consuming. In that case we use the step-deviation method in which
we take the deviations of the given values of X from any arbitrary point A. Generally, ‘A’ is taken to be a
value lying in the middle part of the distribution, although the formula (6.25) holds for any value of A.
(6·25) leads to the following important conclusion :
“The variance and consequently the standard deviation of a distribution is independent of the change
of origin”.
Thus, if we add (subtract) a constant to (from) each observation of the series, its variance remains
same.
Mathematically this means that :
Var (X + a) = Var (X – b) = Var X, where a and b are constants.
…(6.25a)
Formula 3. If we change the origin and scale in X i.e., if we take :
X – A;
d =
h > 0, then
h
σx2 = h2 σd 2 = h2
[ N1 ∑ fd – (N1 (∑ fd) ) ]
2
2
…(6·26)
In case of grouped or continuous frequency distribution, it is convenient to change the scale also. Thus,
if h is the magnitude of the class interval (or if h is the common factor in the values of the variable X), then
we may take
X – A,
d =
h>0
and
use (6·26)
h
Remarks 1. Formula (6.26) also leads to the following result.
Var (aX) = a2 Var (X)
⇒
S.D. (aX) = a S.D. (X)
…(6·26a)
where Var (X) denotes the variance of X and a is a constant.
This shows that variance (or s.d.) is not independent of change of scale.
Combining the results obtained in (6·25) and (6·26) we conclude that :
“Variance or standard deviation is independent of the change of origin but not of the scale”.
2. For numerical problems, a somewhat more convenient form of (6·26) may be used. Rewriting (6·26),
we get
1/2
h2
h
σx2 = 2 N ∑ f d2 – (∑ f d)2 ⇒ σx =
N ∑ f d 2 – (∑ f d)2
⇒ σ x = h . σd
…(6·27)
N
N
[
]
[
]
3. Different Formulae for Variance for Raw Data
If x 1 , x2,…, xn are the n observations, then
—
1
σx2 = ∑ (X – X )2
n
⇒
—
1
σx2 = ∑ X2 – X
n
—
2
=
1
∑X
n
2
–
( 1n ∑ X )
…(6·28)
2
[Taking f = 1 and N = n in (6·24)]
…(6·28a)
4. If we are given X and σx2, then we can obtain the values of ∑ X and ∑ X 2 as discussed below.
From (6·28a), we have
—
—
1
X = ∑X
⇒
∑ X = nX
…(6·28b)
n
–
–
1
and
σx2 = ∑ X2 – X 2
⇒
∑ X2 = n (σx2 + X 2 )
…(6·28c)
n
Formulae (6·28b) and (6·28c) are very useful when we are given the values of the mean and standard
deviation (or variance) and later on it is found that one or more of the observations are wrong and it is
6·21
MEASURES OF DISPERSION
required to compute the mean and variance after replacing the wrong values by correct values or after
deleting the wrong values. For illustrations, see Examples 6·24 to 6·26.
Example 6·10. Calculate the standard deviation of the following observations on a certain variable :
240·12,
240·13,
240·15,
240·12,
240·17,
240·15,
240·17,
240·16,
240·22,
240·21
Solution.
COMPUTATION OF STANDARD DEVIATION
∑X 2401·60
X =
= 10 = 240·16
10
—
1
σ2 = ∑(X – X )2 = 0·0106
10 = 0·00106
n
—
X
240·12
240·13
240·15
240·12
240·17
240·15
240·17
240·16
240·22
240·21
1
∴ s.d. (σ) = (0·00106)2
[
= Antilog [
= Antilog [
= Antilog [
= Antilog
1
2
log (0·00106)
–
1
2
(3·0253)
]
1
2
(–3 + 0·0253)
1
2
(–2·9747)
]
—
—
(X – X )2
X –X
—————————————————————
—————————————————————
———————————————————————————
]
]
– 0·04
– 0·03
– 0·01
– 0·04
0·01
– 0·01
0·01
0·00
0·06
0·05
0·0016
0·0009
0·0001
0·0016
0·0001
0·0001
0·0001
0
0·0036
0·0025
———————————————————— ———————————————————————————————————————————————
–
= Antilog (–1·4173) = Antilog (2·5127)
= 0·03256
∑X = 2401·60
—
—
∑(X – X ) = 0 ∑ (X – X )2 = 0·0106
Example 6·11. Complete a table showing the frequencies with which words of different numbers of
letters occur in the extract reproduced below (omitting punctuation marks) treating as the variable the
number of letters in each word, and obtain the mean and standard deviation of the distribution :
“Her eyes were blue : blue as autumn distance — blue as the blue we see, between the retreating
mouldings of hills and woody slopes on a sunny September morning : a misty and shady blue, that had no
beginning or surface, and was looked into rather than at.”
Solution. Here we take the variable (X) as the number of letters in each word in the extract given
above. We find that in the extract given above there are words with number of letters ranging from 1 to 10.
Hence, the variable X takes the values from 1 to 10. The frequency distribution is easily obtained by using
‘tally marks’ as given in the following Table.
Mean = A +
∑ fd
=6+
N
( )
–76
46
= 6 – 1·65 = 4·35
σ2 =
=
⇒
1
∑ fd 2 –
N
354
46 –
( N1 ∑ fd )
2
(–1·65)2
= 7·6956 – 2·7225 = 4·9731
σ = ⎯√⎯⎯⎯⎯
4·9731 = 2·23.
No. of letters in
Frequency (f)
a word (X)
d=X–A
=X–6
fd
fd2
————————————————— ————————————— ———————————— ————————————
—————————————
1
2
3
4
5
6
7
8
9
10
Total
2
8
9
10
5
4
3
1
3
1
N = ∑ f = 46
–5
–4
–3
–2
–1
0
1
2
3
4
–10
–32
–27
–20
–5
0
3
2
9
4
50
128
81
40
5
0
3
4
27
16
∑ fd = –76 ∑ fd2 = 354
Example 6·12. Calculate the mean and standard deviation from the following data :
Value
:
90—99
80—89
70—79
60—69
50—59
40—49
Frequency :
2
12
22
20
14
4
30—39
1
6·22
BUSINESS STATISTICS
Solution.
CALCULATIONS FOR MEAN AND S.D.
x – 64·5
10
fd
fd2
3
6
18
12
2
24
48
22
1
22
22
Class
Mid-value
(x)
Frequency
(f)
90—99
94·5
2
80—89
84·5
70—79
74·5
60—69
64·5
20
0
0
0
50—59
54·5
14
–1
–14
14
40—49
44·5
4
–2
–8
16
30—39
34·5
1
–3
–3
9
∑ fd = 27
∑fd = 127
Total
d=
N = 75
Mean = A +
2
h∑ fd
10 × 27
= 64·5 +
= 64·5 + 3·6 = 68·1
N
75
∑ fd2
–
N
S.D. = h .
( ∑Nfd )
2
= 10 ×
127
75 –
( )
27 2
75
= 10 × √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
1·6933 – 0·1296
= 10 × √⎯⎯⎯⎯⎯
1·5637 = 10 × 1·2505 = 12·505.
Example 6·13. The arithmetic mean and the standard deviation of a set of 9 items are 43 and 5
respectively. If an item of value 63 is added to the set, find the mean and standard deviation of 10 items.
[Delhi Univ. B.A. (Econ. Hons.), 2006]
Solution. We are given : n = 9, –x = 43 and σ = 5.
–x = ∑x
⇒
∑ x = nx– = 9 × 43 = 387
n
∑x2
Also
σ2 =
– ( –x )2 ⇒
∑ x2 = n(σ2 + –x 2 )
n
∴
∑x 2 = 9(25 + 432 ) = 9(25 + 1849) = 9 × 1874 = 16866
If a new item 63 is added, then number of items becomes 10.
New (∑x) = ∑x + 63 = 387 + 63 = 450
∴
New mean =
New
(∑x2)
450
10 =
=∑x2
New s.d. =
…(i)
…(ii)
[From (i)]
45
+ 63 2 = 16866 + 3969 = 20835
New (∑x2)
– (New mean)2 =
10
20835
2
10 – (45)
= ⎯√⎯⎯⎯⎯⎯⎯⎯⎯
2083·5 – 2025 = ⎯
√⎯⎯⎯
58·5 = 7·65.
Example 6·14. Twenty passengers were found ticketless on a bus. The sum of squares and the S.D. of
the amount found in their pockets were Rs. 2,000·00 and Rs. 6·00 respectively. If the total fine imposed on
these passengers is equal to the total amount recovered from them and fine imposed is uniform, what is the
amount each one of them has to pay as fine ? What difficulties do you visualize if such a system of penalty
were imposed ?
[Delhi Univ. B.A. (Econ.)Hons., 1993]
Solution. Let xi, i = 1, 2,…,20 be the amount (in Rs.) found in the pocket of the ith passenger. Then we
are given :
20
n = 20,
∑ xi2 = Rs. 2,000
i=1
and
s.d.(σ) = Rs. 6
…(i)
6·23
MEASURES OF DISPERSION
The total fine imposed on the ticketless passengers is given to be equal to the total amount recovered
from them.
20
∴
Total fine imposed on the 20 passengers =
∑ xi.
i=1
Further, since the fine imposed is uniform among all the 20 passengers,
n
1
Fine to be paid by each passenger = 20
∴
1
We have : σ2 = ∑ xi2 – –x 2
n
–x 2 = Rs.2
∴
(
2000
20
–x 2
⇒
=
∑
i=1
1
20
xi = –x
… (ii)
∑ x i 2 – σ2
)
– 62 = Rs.2 (100 – 36) = Rs.2 64, ⇒
–x = Rs. 8
[From (i)]
Hence, using (ii), the fine paid by each of the passengers is Rs. 8.
If among these ticketless passengers, there are a few rich persons with large sums of money in their
pockets, then an obvious shortcoming of this system of imposing penalty is that, it will give undue heavy
penalty to the poor passengers (with smaller amounts of money in their pockets).
1
Example 6.15. The variance of a series of numbers 2, 3, 11 and x is 12 . Find the value of x.
4
[I.C.W.A. (Foundation), June 2006]
Solution. We are given n = 4.
X
2
3
11
x ∑X = 16 + x
∑X2
∑X 2 x2 + 134
16 + x 2
2
σX2 =
–
=
–
X
4
9
121 x2 ∑X2 = x2 + 134
n
n
4
4
( )
(
)
1
1
[4 (x2 + 134) – (16 + x)2 ] = 16
[4x2 + 536 – (256 + x2 + 32x)]
16
1
=
[3x2 – 32x + 280] = 12 41 = 49
(Given).
16
4
3x2 – 32x + 280 = 49 × 4 = 196
⇒
3x 2 – 32x + 84 = 0
=
⇒
⇒
σX2
∴
x =
⇒
x =
32 ± √
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
(– 32)2 – 4 × 3 × 84 32 ± √
⎯⎯⎯⎯⎯⎯⎯⎯
1024 – 1008 32 ± 4
=
=
2×3
6
6
(
32 + 4
6
or
) (
32 – 4
= 6
6
or
)
14
.
3
Example 6·16. The mean of 5 observations is 4·4 and the variance is 8·24. If three of the five
observations are 1, 2 and 6, find the values of the other two.
[Delhi Univ. B.A. (Econ. Hons.), 2004]
–
Solution. We are given n = 5, x = 4·4 and σ2 = 8·24
We have :
∑x = nx– = 5 × 4·4 = 22
∑x 2 = n (σ2 + –x 2 ) = 5(8·24 + 19·36) = 5 × 27·60 = 138
and
Three observations are 1, 2 and 6. Let the two unknown observations be x 1 and x2 . Then
∑x = 1 + 2 + 6 + x1 + x2 = 22
⇒
∑x 2 = 1 2 + 2 2 + 6 2 + x12 + x22 = 138 ⇒
Substituting the value of x2 from (*) in (**) we get
x12 + (13 – x1)2 = 97
2
2
2
⇒
x1 + [13 + x1 – 2 × 13 × x1 ] = 97
x1 + x2 = 22 – 9 = 13
x12 + x22 = 138 – 41 = 97
[·. · (a – b)2 = a 2 + b2 – 2ab]
…(*)
…(**)
6·24
BUSINESS STATISTICS
⇒
x12 + (169 + x 1 2 – 26x1) = 97
⇒
2x12 – 26x1 + 72 = 0
Solving as a quadratic equation in x1 , we get
26 ± √
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
(26) 2 – 4 × 2 × 72 26 ± √
⎯⎯⎯⎯⎯⎯⎯
676 – 576 26 ± 10
=
=
2×2
4
4
26 + 10
26 – 10
=
or
= 9 or 4
4
4
Substituting in (*), we get
x1 = 9
⇒ x2 = 13 – 9 = 4 or x 1 = 4
Hence, the other two numbers are 4 and 9.
Example 6.17. (a) For a group of 10 items,
x1 =
10
10
i=1
i=1
∑ (Xi – 2) = 40 and
x2 = 13 – 4 = 9
∑ Xi2 = 495.
Then find the variance of this group.
(b) For 10 values X1 , X2 , …, X10 of the variable X,
10
10
i=1
i=1
∑ Xi = 110 and
⇒
[I.C.W.A. (Foundation), June 2005]
∑ (Xi – 5) 2 = 1,000.
Find variance of X.
[I.C.W.A. (Foundation), June 2006]
Solution. (a) In the usual notations, we are given :
n = 10 ; ∑X2 = 495 and
∑ (X – 2) = 40
⇒
∑X – 2n = 40
⇒
∑X = 40 + 2 × 10 = 60
2
2
2
∑X
∑X
495
60
∴
σX2 =
–
=
–
= 49.5 – 36 = 13.5
n
n
10
10
(b) We are given : n = 10, ∑X = 110 and ∑ (X – 5)2 = 1,000.
Let d = X – A = X – 5, (A = 5). Then, we have
∑d = ∑ (X – 5) = ∑X – 5 × n = 110 – 5 × 10 = 60 and ∑ d 2 = ∑ (X – 5)2 = 1,000.
Since variance is independent of change of origin, we get :
( )
( )
( )
( )
∑d2
∑d 2 1‚000
60 2
–
=
–
= 100 – 36 = 64.
n
n
10
10
Example 6.18. For the numbers 5, 6, 7, 8, 10, 12, if S1 and S 2 are the respective Root Mean Square
Deviations about the mean and about an arbitrary number 9, show that 17S2 2 = 20 S12 .
[I.C.W.A. (Foundation), June 2003]
σX2 = σd 2 =
Solution. We have
CALCULATIONS FOR MEAN SQUARE DEVIATIONS
—
—
— ∑X 48
X
(X – 9)
(X – 9)2
X–X
(X – X )2
n = 6,
X=
=
=8
n
6
=X–8
16
–4
9
–3
5
S1 2 = Mean Square Deviation about mean
9
–3
4
–2
6
— 2 34 17
1
= ∑ (X – X ) = =
4
–2
1
–1
7
n
6
3
1
–
1
0
0
8
S2 2 = Mean Square Deviation about the
1
1
4
2
10
point ‘9’
9
3
16
4
12
—
1
40
20
∑(X
– 9)2
∑X
=
48
2
∑(X – X )
= ∑ (X – 9)2 = =
n
6
3
= 40
= 34
∴
2
S1 17 3 17
=
× =
S2 2 3 20 20
⇒
17 S2 2 = 20 S1 2
6·25
MEASURES OF DISPERSION
Example 6·19. A charitable organisation decided to give old age pensions to people over sixty years of
age. The scale of pensions were fixed as follow :
Age group
60—65
Rs. 200 per month
”
65—70
Rs. 250 per month
”
70—75
Rs. 300 per month
”
75—80
Rs. 350 per month
”
80—85
Rs. 400 per month
The ages of 25 persons who secured the pensions right are as given below :
74,
62,
84,
72,
61,
83,
72,
81,
64,
71,
63,
61,
60,
67,
74,
64,
79,
73,
75,
76,
69,
68,
78,
66,
67
Calculate monthly average pensions payable per person and the standard deviation.
[Delhi Univ. B.Com. (Hons.) (External), 2005]
Solution. First of all we shall prepare the frequency distribution of the 25 persons with respect to age
in the age-groups 60—65, 65—70,…, 80—85 (as suggested by the above data) by using the method of tally
marks. Then we shall compute the arithmetic mean of the pension payable per person and also its standard
deviation, as explained in the following table.
COMPUTATION OF MEAN AND STANDARD DEVIATION
Age Group
60—65
65—70
70—75
75—80
80—85
Tally makrs
||||
||||
||||
Frequency (f)
Monthly pension
(in Rs.)
200
250
300
350
400
7
5
6
4
3
||
|
||||
|||
d=
X – 300
50
–2
–1
0
1
2
fd
–14
–5
0
4
6
∑ fd = –9
N = 25
fd2
28
5
0
4
12
∑ fd2 = 49
Average monthly pension is given by :
—
h∑fd
X =A+
= 300 + 50 ×25(–9) = (300 – 18) = Rs. 282
N
Standard deviation of monthly pension is :
σ =h.
( )
∑ fd2 ∑ fd
–
N
N
2
=
h
N
50
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
N ∑fd2 – (∑ fd)2 = 25
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
25 × 49 – (–9) 2
= 2⎯
√⎯⎯⎯⎯⎯⎯
1225 – 81 = 2⎯
√⎯⎯⎯
1144 = 2 Antilog
= 2 Antilog
(
1
2
× 3·0585
(
1
2
log 1144
)
) = 2 Antilog (1·5292) = 2 × 33·83 = Rs. 67·66.
Example 6·20. You are the incharge of the rationing department of a state affected by food shortage.
The following information is received from your local investigators :
Area
X
Y
Mean Calories
2,500
2,200
Standard Deviation of Calories
500
300
The estimated requirement of an adult is taken at 3,000 calories daily and absolute minimum at
1,250. Comment on the reported figures and determine which area needs more urgent action.
[Delhi Univ. B.Com. (Hons.), 2002]
–
Solution. We shall compute the 3-sigma limits x ± 3σ for each area, which will include approximately
99·73% of the population observation [assuming that the distribution is approximately normal].
6·26
BUSINESS STATISTICS
—
3-σ Limits = X ± 3σ
2500 ± 3 × 500 = 2500 ± 1500 = (1000, 4000)
2200 ± 3 × 300 = 2200 ± 900 = (1300, 3100)
Area X
Area Y
The absolute daily minimum calories requirement for a person is 1250. From the above figures we
observe that almost all the persons in the area Y are getting more than the minimum calories requirement as
the lower limit in this area is 1300. However, since in the area X, the lower 3-σ limit is 1000 which is less
than 1250, quite a number of people in area X are not getting the minimum requirement of 1250 calories.
Hence, as the incharge of the rationing department, it becomes my duty to take urgent action for the people
of area X.
Example 6·21. Find the proportion of items lying within : (i) mean ± σ and
(ii) mean ± 2σ,
of the following distribution.
Class
Frequency
11—12
13—14
15—16
17—18
19—20
Class
5
426
720
741
665
Frequency
21—22
23—24
25—26
27—28
29—30
395
38
8
5
7
Solution.
COMPUTATION OF MEAN AND S.D.
Class
interval
Mid-value
(X)
11—12
13—14
15—16
17—18
19—20
21—22
23—24
25—26
27—28
29—30
11·5
13·5
15·5
17·5
19·5
21·5
23·5
25·5
27·5
29·5
d=
X – 19·5
2
–4
–3
–2
–1
0
1
2
3
4
5
f
fd
fd 2
5
426
720
741
665
395
38
8
5
7
–20
–1278
–1440
–741
0
395
76
24
20
35
80
3834
2880
741
0
395
152
72
80
175
∑ f = 3010
Mean = A +
σ =h
∑ fd = –2929
∑ fd 2 = 8409
h ∑fd
(–2929)
= 19·5 + 2 × 3010
= 19·5 – 1·946 = 17·55
N
∑ fd2
–
N
( ∑Nfd )
2
=2×
8409
3010
–
(
)
–2929 2
3010
= 2 × √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
2·7937 – (0·9731)2 = 2 × ⎯
√⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
2·7937 – 0·9469
= 2 × √⎯⎯⎯⎯⎯
1·8464 = 2 × 1·35897 = 2·7179 ≈ 2·72
∴
Mean ± σ = 17·55 ± 2·72 = 20·27 and 14·83
and
Mean ± 2σ = 17·55 ± 2 × 2·72 = 17·55 ± 5·44 = 22·99 and 12·11.
The number of items lying within Mean ± σ i.e., within 14·83 and 20·27 is 720 + 741 + 665 = 2126,
and the proportion of items lying within Mean ± σ is :
2‚126
3‚010 =
0·7063
i.e.,
70·63%
6·27
MEASURES OF DISPERSION
The number of items lying within Mean ± 2σ i.e., within 12·11 and 22·99 is : 426 + 720 + 741 + 665
+ 395 = 2,947. Hence, the proportion of items lying within Mean ± 2σ is :
2‚947
3‚010 =
0·9791
i.e.,
97·91%
Example 6·22. The mean and standard deviation of the frequency distribution of a continuous random
variable X are 40·604 lbs. and 7·92 lbs. respectively. The distribution after change of origin and scale is as
follows :
d
:
–3
–2
–1
0
1
2
3
4
Total
f
:
3
15
45
57
50
36
25
9
240
where d = (X – A)/h and f is the frequency of X. Determine the actual class intervals.
Solution.
COMPUTATION OF MEAN AND S.D.
We are given : d = (X – A)/h
d
X = 40·604
—
X =A+
⇒
⇒
and
σx = 7·92
h ∑ fd
N
40·604 = A + h
( )
149
240
40·604 = A + 0·621h
σx = h
⇒
f
fd
fd2
———————————————
——————————————————————
——————————————————
——————————————————
—
7·92 = h
…(*)
( )
∑ fd2 ∑ fd
–
N
N
695
240
–
( )
149 2
240
2
–3
3
–9
27
–2
15
–30
60
–1
45
– 45
45
0
57
0
0
1
50
50
50
2
36
72
144
3
25
75
225
4
9
36
144
N = ∑ f = 240
∑ fd = 149
—————————————————————————————————————
———————————————————
———————————————————
∑ fd2 = 695
= h⎯
√⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
2·8958 – (0·6208) 2
=h√
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
2·8958 – 0·3854 = h ⎯
√⎯⎯⎯⎯
2·5104 = 1·5844 h
⇒
7·92
h = 1·5844 = 4·9987 ~– 5
Substituting in (*), we get
A = 40·604 – 5 × 0·621
= 40·604 – 3·105 = 37·499 ~– 37·5
The value
d=0
⇒
X–A=0
⇒ X = A = 37·5
Since the magnitude of the class interval is h = 5,
the boundaries of the corresponding class are
(37·5 – 2·5, 37·5 + 2·5) i.e., (35, 40). Thus the actual
frequency distribution is as given in the adjoining table.
d
X = Mid-value
of Class
Class Interval
22·5
27·5
32·5
37·5
42·5
47·5
52·5
57·5
20—25
25—30
30—35
35—40
40—45
45—50
50—55
55—60
Frequency
————————————
———————————————————————————————————————————
————————————————
–3
–2
–1
0
1
2
3
4
3
15
45
57
50
36
25
9
Example 6·23. (a) Find the mean and standard deviation of the first n natural numbers.
(b) Hence deduce the mean and s.d. of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Solution. (a) If the variable X denotes the natural number, then the first n natural numbers are
1, 2, 3, …, n.
X
1
2
3
…
n
2
2
2
2
X
1
2
3
…
n2
6·28
BUSINESS STATISTICS
∑X
1 + 2 + 3 + … + n n(n + 1) . 1
n +1
=
=
=
n
n
2
n
2
∑X2 —2 12 + 2 2 + 3 3 + … + n2
n+1 2
Variance =
–X =
–
n
n
2
2
n(n + 1) (2n + 1)
(n + 1) (n + 1)
=
–
=
2(2n + 1) – 3(n + 1)
6n
4
12
(n + 1)
(n + 1) (n – 1)
=
4n + 2 – 3n – 3 =
12
12
Mean =
(
)
[
[
∴
σ2 =
…(*)
]
]
n2 – 1
12
⇒
s.d. (σ) =
n2 – 1
12
… (**)
Note. This is a standard result and should be committed to memory.
(b) Deduction. We have to find mean and s.d. of the first 10 natural numbers. Taking n = 10 in (*) and
(**), we get respectively :
Mean =
10 + 1 11
=
= 5·5 ;
2
2
s.d. (σ) =
102 – 1
12
=
100 – 1
12
=√
⎯⎯⎯
8·25 = 2·87.
Example 6·24. The arithmetic mean and standard deviation of series of 20 items were calculated by a
student as 20 cm. and 5 cm. respectively. But while calculating them an item 13 was misread as 30. Find
the correct arithmetic mean and standard deviation.
—
Solution. We are given n = 20, X = 20 cms, σ = 5 cms ; Wrong value used = 30 ; Correct value = 13
—
—
We have : ∑X = nX = 20 × 20 = 400
and
∑X2 = n(σ2 + X 2 ) = 20(25 + 400) = 8500
If the wrong observation 30 is replaced by the correct value 13, then the number of observations
remains same viz., 20 and
Corrected ∑X = 400 – 30 + 13 = 383
; Corrected ∑X2 = 8500 – (30)2 + (13)2 = 7769
Corrected (∑X) 383
∴
Corrected mean =
= 20 = 19·15
n
Corrected ( ∑X )2 –
Corrected σ2 =
(Corrected mean)2
n
7769
=
– (19·15)2 = 388·45 – 366·72 = 21·73
20
∴
Corrected s.d. (σ) = ⎯√⎯⎯⎯
21·73 = 4·6615.
Example 6·25. The mean and the variance of ten observations are known to be 17 and 33
respectively. Later it is found that one observation (i.e., 26) is inaccurate and is removed. What is the mean
and standard deviation of the remaining ?
[Delhi Univ. B.A. (Econ. Hons.), 2009; C.A. (Foundation), May 2002]
—
Solution. In the usual notations, we are given : n = 10, X = 17 and σ2 = 33
—
—
∑X = nX = 10 × 17 = 170 and ∑X2 = n(σ2 + X 2 ) = 10 (33 + 289) = 3220
If the inaccurate observation (26) is removed, then for the remaining n – 1 = 10 – 1 = 9 observations,
the values of ∑X and∑X2 are given by :
Corrected ∑X = (∑X) – 26 = 170 – 26 = 144
Corrected ∑X2 = (∑X2 ) – 262 = 3220 – 676 = 2544
—
The mean ( X 1 ) and variance (σ12 ) of the remaining 9 observations are given by :
—
Corrected ∑X 144
New mean ( X 1 ) =
= 9 = 16
9
Corrected ∑X2 — 2 2544
New variance (σ1 2 ) =
– (X 1 ) = 9 – 256 = 282·67 – 256 = 26·67.
9
6·29
MEASURES OF DISPERSION
Example 6·26. For a frequency distribution of marks in Sociology of 200 candidates (grouped in
intervals 0—5, 5—10,…,etc.), the mean and the standard deviation was found to be 45 and 15. Later it was
discovered that the score 53 was misread as 63 in obtaining the frequency distribution. Find the correct
mean and standard deviation corresponding to the correct frequency distribution.
—
Solution. We are given N = 200, X = 45 and σ = 15. These values have been obtained on using the
wrong value 63 while the correct value is 53. In case of grouped or continuous frequency distribution, the
value of X used for computing the mean and the standard deviation, is the mid-value of the class interval.
Since the wrong value 63 lies in the interval 60—65 with the mid-value 62·5 and the correct value 53 lies in
—
the interval 50—55 with mid-value 52·5, the question amounts to finding the correct values of X and σ if
the wrong value 62·5 is replaced by the correct value 52·5.
—
∴ Corrected ∑ f X = 9000 – 62·5 + 52·5 = 8990
∴ We have N = 200, X = 45, σ = 15
Corrected ∑ f X2 = 450000 – (62·5)2 + (52·5)2
Wrong value used = 62·5 ; Correct value = 52·5
—
= 450000 – [(62·5) 2 – (52·5)2 ]
∑ f X = NX = 200 × 45 = 9000
= 450000 – (62·5 + 52·5) (62·5 – 52·5)
—
∑ f X2 = N(σ2 + X2 ) = 200 (152 + 45 2 )
= 450000 – 115 × 10 = 450000 – 1150
= 200(225 + 2025) = 200 × 2250 = 450000
= 448850
Corrected ∑ f X 8990
∴
Corrected mean =
= 200 = 44·95
N
Corrected σ =
=
Corrected ∑ f X2
– (Corrected mean)2
N
448850
200 –
(44·95)2 = √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
2244·25 – 2020·50 = √
⎯⎯⎯⎯⎯⎯
223·75 = 14·96.
EXERCISE 6·3
1. (a) Explain with suitable example the term variation. What purposes does a measure of variation serve ?
Comment on some of the well-known measures of variation along with their respective merits and demerits.
[Delhi Univ. MBA, 2000]
(b) What is a measure of dispersion ? Discuss four important measures of spread indicating their uses.
[Andhra Pradesh Univ. B.Com., 1998]
2. What is meant by dispersion ? In your opinion which is the best method of finding out dispersion and why ?
[Delhi Univ. B.Com. (Pass), 1999]
3. What are the chief requisites of a good measure of dispersion ? In the light of those, comment on some of the
well-known measures of dispersion.
4. (a) What do you understand by absolute and relative measures of dispersion ? Explain advantages of the relative
measures over the absolute measures of dispersion.
(b) Define mean deviation and standard deviation. Explain, why economists prefer mean deviation to standard
deviation in their analysis.
5. (a) Compare mean deviation and standard deviation as measures of variation. Which of the two is a better
measure ? Why ?
[Delhi Univ. B.Com. (Hons.), 2001]
(b) Explain the mathematical properties of standard deviation. Why is standard deviation used more than mean
deviation ?
[Delhi Univ. B.Com. (Hons.), 2009]
6. What is standard deviation ? Explain its superiority over other measures of dispersion.
7. Give the various formulae for computing the standard deviation.
8. State giving reasons whether the following statements are true or false :
(i) Standard deviation can never be negative.
(ii) The sum of squared deviations measured from mean is least.
Ans. (i) True, (ii) True.
9. For the numbers (X) : 1, 3, 4, 5 and 12, find :
(i) the value (v) for which ∑ (X – v)2 is minimized.
(ii) the value (v) for which ∑ | X – v | is minimized.
[Delhi Univ. B.A. (Econ. Hons.), 2009]
6·30
BUSINESS STATISTICS
—
1
Ans. (i) ∑ (X – v)2 is minimum when v = X = 5 (1 + 3 + 4 + 5 + 12) = 5
(ii) ∑ | X – v | is minimum when v = Median of (1, 3, 4, 5, 12) = 4
10. Calculate standard deviation of the following marks obtained by 5 students in a tutorial group :
Marks Obtained : 8, 12, 13, 15, 22
[Delhi Univ. B.Com. (Pass), 1997]
Ans. σ2 =
( )
∑x2
∑x
–
n
n
2
=
( )
1086
70
–
5
5
2
= 21·2
⇒
σ =⎯
√⎯⎯
21·2 = 4·6.
11. Why is standard deviation considered to be the best measure of dispersion ? Find the variance if ∑d2 = 150 and
N = 6. Deviations are taken from actual mean.
[Delhi Univ. B.Com. (Pass), 1998]
150
2
Ans. σ =
= 25.
6
12. (a) From the following information, find the standard deviation of x and y variables :
∑x = 235, ∑y = 250, ∑x2 = 6750, ∑y2 = 6840, N = 10.
[Delhi Univ. B.Com. (Hons.) 1997]
Ans. σ x = 11·08; σy = 7·68.
(b) You are given the following raw sums in a statistical survey of two variables X and Y :
∑X = 240, ∑Y = 250, ∑X2 = 6400 and ∑Y2 = 7060.
Ten items are included in each survey. Compute Standard Deviation of the X and Y variables.
[Delhi Univ. B.Com. (Pass), 1996]
Ans. σx = 8, σy = 9.
13. (a) State a formula for computing standard deviation of n natural numbers 1, 2, …, n.
[Delhi Univ. B.Com. (Pass), 2000]
Ans. σ = √
⎯⎯⎯⎯⎯⎯⎯
(n2 – 1) / 12.
⎯ 2 . [Kerala Univ. B.Com., 1996]
(b) Show that the standard deviation of the natural numbers 1, 2, 3, 4 and 5 is √
(c) Mean of 10 items is 50 and S.D. is 14. Find the sum of the squares of all the items.
[Mahatma Gandhi Univ. B.Com., April 1998]
Ans. ∑x2 = 26960
14. Calculate standard deviation of the following series.
Daily Wages of
No. of
Daily Wages of
No. of
Daily Wages of
No. of
Workers (in Rs.)
Workers
Workers (in Rs.)
Workers
Workers (in Rs.)
Workers
100—105
200
120—125
350
140—145
280
105—110
210
125—130
520
145—150
210
110—115
230
130—135
410
150—155
160
115—120
320
135—140
320
155—160
90
Ans. s.d. = 14.244
15. Find out the mean and standard deviation of the following data.
Age under (years)
:
10
20
30
40
50
60
70
80
No. of persons dying
:
15
30
53
75
100
110
115
125
Ans. Mean = 35.16 years, S.D. = 19.76 years.
16. In the following data, two class frequencies are missing.
Class Interval
Frequency
Class Interval
100—110
4
150—160
110—120
7
160—170
120—130
15
170—180
130—140
—
180—190
140—150
40
190—200
Frequency
—
16
10
6
3
However, it was possible to ascertain that the total number of frequencies was 150 and that the median has been
correctly found out as 146·25. You are required to find with the help of information given :
(i) The two missing frequencies.
(ii) Having found the missing frequencies, calculate arithmetic mean and standard deviation.
(iii) Without using the direct formula, find the value of mode.
6·31
MEASURES OF DISPERSION
Ans. (i) 24, 25; (ii) A.M. = 147·33, s.d. = 19·2; (iii) Mode = 144·09.
17. The following table gives the distribution of income of households based on hypothetical data :
Income
Percentage of
Income
Percentage of
households
(Rs.)
(Rs.)
households
Under 100
7·2
500—599
14·9
100—199
11·7
600—699
10·4
200—299
12·1
700—999
9·0
300—399
14·8
1000 and above
4·0
400—499
15·9
(i) What are the problems involved in computing standard deviation from the above data ?
(ii) Compute a suitable measure of dispersion.
Ans. (ii) Compute Quartile Deviation. Q.D. = 169·425 ; Coeff. of Q.D. = 0·404.
18. The standard deviation calculated from a set of 32 observations is 5. If the sum of the observations is 80, what
is the sum of the squares of these observations ?
Ans. ∑X2 = 1000.
19. The mean of 200 items is 48 and their standard deviation is 3. Find the sum and sum of squares of all items.
Ans. 9,600 ; 4,62,600.
—
20. Given : No. of observations (N) = 100; Arithmetic average (X ) = 2 ; Standard deviation (sx) = 4
find ∑X and ∑X2.
Ans. ∑X = 200, ∑X2 = 2000.
21. The mean of 5 observations is 3 and variance is 2. If three of the five observations are 1, 3, 5, find the other
two.
Ans. 2, 4.
22. An association doing charity work decided to give old age pension to people of 60 years and above in age. The
scales of pension were fixed as follows :
Age group 60—65 ; Rs. 400 per month
Age group 65—70 ; Rs. 500 per month
Age group 70—75 ; Rs. 600 per month
Age group 75—80 ; Rs. 700 per month
Age group 80—85 ; Rs. 800 per month
85 and above
; Rs. 1000 per month.
The ages of 30 persons who secured the pension right are given below :
62
65
68
72
75
77
82
85
90
78
75
61
60
68
72
76
78
79
80
82
68
75
94
98
73
77
68
65
71
89
Calculate the monthly average pension payable and the standard deviation .
Ans. Average monthly pension = Rs. 676·70; s.d. = Rs. 183·80.
23. Treating the number of letters in each word in the following passage as the variable X, prepare the frequency
distribution table and obtain its mean, median, mode and standard deviation.
“The reliability of data must always be examined before any attempt is made to base conclusions upon them. This
is true of all data, but particularly so of numerical data, which do not carry their quality written large on them. It is a
waste of time to apply the refined theoretical methods of Statistics to data which are suspect from the beginning.”
Ans. Mean = 4·565, Median = 4, Mode = 3, S.D. = 2·673.
24. A collar manufacturer is considering the production of a new style of collar to attract young men. The
following statistics of new circumferences are available based on measurements of a typical group :
Mid-value (in inches)
No. of students
Mid-value (in inches)
No. of students
12·0
4
14·5
18
12·5
8
15·0
10
13·0
13
15·5
7
13·5
20
16·0
5
6·32
BUSINESS STATISTICS
14·0
25
—
Use the criterion X ± 3σ to obtain the largest and the smallest size of collar he should make in order to meet the
needs of practically all his customers having in mind that collars are worn on an average 3/4 inches larger than neck
size.
Hint. Mean = 13·968″; S.D. = 0·964″. ; Limits for collar size are given by : [Mean ± 3 s.d.] + 3/4 ;
Largest collar size = 14·718 + 2·892 = 17·61″ ; Smallest collar size = 14·718 – 2·892 = 11·826″
25. The following data represent the percentage impurities in a certain chemical substance.
Percentage of impurities
Frequency
Percentage of impurities
Frequency
Less than 5
0
10—10·9
45
5—5·9
1
11—11·9
30
6—6·9
6
12—12·9
5
7—7·9
29
13—13·9
3
8—8·9
75
14—14·9
1
9—9·9
85
(i) Calculate the mean and standard deviation.
(ii) Find the number of frequency lying between (A.M. ± 2 S.D.).
Ans. (i) Mean = 9·3857, S.D. = 1·3924;
(ii) 267.
26. The following distribution was obtained by a change of origin and scale of variable X :
d :
–4
–3
–2
–1
0
1
2
f :
4
8
14
18
20
14
10
3
6
4
6
Write down the frequency distribution of X if it is given that mean and variance are 59·5 and 413 respectively.
Ans.
C.I.
f
C.I.
f
C.I.
f
15·5—25·5
4
45·5—55·5
18
75·5—85·5
10
25·5—35·5
8
55·5—65·5
20
85·5—95·5
6
35·5—45·5
14
65·5—75·5
14
95·5—105·5
6
—————————
Total
100
27. Mean and standard deviation of the following continuous series are 31 and 15·9 respectively. The distribution
after taking step deviation is as follows :
d
:
–3
–2
–1
0
1
2
3
f
:
10
15
25
25
10
10
5
Determine the actual class intervals.
[G.G.I.P. Univ., B.B.A., May 2004]
Ans. 0—10, 10—20, 20—30, 30—40, 40—50, 50—60, 60—70.
28. (a) The mean and standard deviation of a sample of 100 observations were calculated as 40 and 5·1
respectively by a student who took by mistake 50 instead of 40 for one observation. Calculate the correct mean and
standard deviation.
Ans. Corrected mean = 39·9 and s.d. = 5.
(b) For a number of 51 observations, the arithmetic mean and standard deviation are 58·5 and 11 respectively. It
was found after the calculations were made that one of the observations recorded as 15 was incorrect. Find the mean
and standard deviation of the 50 observations if this incorrect observation is omitted.
Ans. Mean = 59·37 and S.D. (σ) = 9·21.
29. The mean and the standard deviation of a sample of size 10 were found to be 9·5 and 2·5 respectively. Later
on, an additional observation became available. This was 15·0 and was included in the original sample. Find the mean
and the standard deviation of the 11 observations.
Ans. Mean = 10, s.d. = 2·86.
30. (a) The mean and standard deviation of 20 items are found to be 10 and 2 respectively. At the time of checking
it was found that one item 8 was incorrect. Calculate the mean and standard deviation if
(i) the wrong item is omitted, and
(ii) it is replaced by 12.
(b) A study of the age of 100 film stars grouped in intervals of 10—12, 12—14,…, etc., revealed the mean age and
standard deviation to be 32·02 and 13·18 respectively. While checking it was discovered that the age 57 was misread as
27. Calculate the correct mean age and standard deviation.
Ans. (a) (i) Mean = 10·1053, s.d. = 1·997; (ii) Mean = 10·2 s.d. = 1·99.
—————————
MEASURES OF DISPERSION
6·33
(b) Mean = 32·32; s.d. =13·402.
31. (a) Mean and Standard Deviation of 100 items are found to be 40 and 10. At the time of calculation two items
are wrongly taken as 30 and 70 instead of 3 and 27. Find the correct mean and correct standard deviation.
Ans. Mean = 39·3 ; S.D. = 10·24
[Delhi Univ. B.Com. (Pass), 2001]
(b) Mean and coefficient of standard deviation of 100 items are found by a student as 50 and 0·1. If at the time of
calculations two items are wrongly taken as 40 and 50 instead of 60 and 30, find the correct mean and standard
deviation.
[Delhi Univ. B.Com. (Hons.), 1996]
–
x
–
σ
Hint. n = 100, x = 50,
= 0 ·1
⇒
σ= =5
⇒
σ 2 = 25.
10
x
Ans. Mean = 50, σ = √
⎯⎯29 = 5·39.
(c) The mean and the standard deviation of a characteristic of 100 items were found to be 60 and 10 respectively.
At the time of calculations, two items were wrongly taken as 5 and 45 instead of 30 and 20. Calculate the corrected
mean and corrected standard deviation.
[Delhi Univ. B.Com. (Hons.), 2009; C.A. (Foundation), June 1993]
Ans. Corrected mean = 60, Corrected S.D. = 9·62
32. Fill in the blanks :
(i) Algebraic sum of deviations is zero from ……
(ii) The sum of absolute deviations is minimum from ……
(iii) Standard deviation is always …… than range.
(iv) Standard deviation is always …… than mean deviation.
(v) The mean and s.d. of 100 observations are 50 and 10 respectively.
The new :
(a) Mean = ……, s.d. = ……, if 2 is added to each observation.
(b) Mean = ……, s.d. = ……, if 3 is subtracted from each observation.
(c) Mean = ……, s.d. = ……, if each observation is multiplied by 5
(d) If 2 is subtracted from each observation and then it is divided by 5.
(vi) Variance is the …… value of mean square deviation.
(vii) If Q1 = 10, Q3 = 40, the coefficient of quartile deviation is ……
(viii) If 25% of the items in a distribution are less than 10 and 25% are more than 40, the quartile deviation is ……
(ix) The median and s.d. of a distribution are 15 and 5 respectively. If each item is increased by 5, the new median
= …… and s.d. = ……
(x) A computer showed that the s.d. of 40 observations ranging from 120 to 150 is 35. The answer is
correct/wrong. Tick right one.
Ans. (i) Arithmetic mean (ii) Median (iii) Less (iv) Greater (v) (a) 52, 10 (b) 47, 10 (c) 250, 50 (d) 9·6, 2
(vi) Minimum
(vii) 0·6
(viii) 15
(ix) 20, 5
(x) Wrong, since s.d. can’t exceed range.
6·10. STANDARD DEVIATION OF THE COMBINED SERIES
As already pointed out, standard deviation is suitable for algebraic manipulations i.e., if we are given
the averages, the sizes and the standard deviations of a number of groups, then we can obtain the standard
deviation of the resultant group obtained on combining the different groups. Thus if
— —
—
σ1 , σ2 , …, σk are the standard deviations; X 1 , X 2 , … , X k are the arithmetic means;
and
n1 , n2 , …, nk are the sizes, of k groups respectively,
then the standard deviation σ of the combined group of size N = n1 + n2 + … + nk is given by the formula
Nσ2 = n1 (σ12 + d12 ) + n2 (σ22 + d22 ) + … + nk(σk2 + dk2 )
… (6·29)
–
–
– –
– –
where
d1 =X1 – X; d2 = X2 – X, … , dk = Xk – X
… (6·29a)
–
–
–
–
n X + n X + … + nkXk
and
X = 1 1 2 2
, is the mean of the combined group.
… (6·29b)
n1 + n2 + … + n k
Thus the standard deviation of the combined group is given by :
6·34
BUSINESS STATISTICS
σ=
[
]
n1 (σ12 + d12 ) + n2 (σ22 + d 2 2 ) + … + nk(σk2 + dk2 ) 1/2
n1 + n2 + … + nk
… (6·30)
In particular, for two groups we get from (6·33) :
(n1 + n2) σ2 = n1 (σ12 + d12 ) + n2 (σ22 + d22 )
–
–
–
–
–
– –
n X + n2X2 n2 (X1 – X2 )
where
d1 = X1 – X = X1 – 1 1
=
n1 + n2
n1 + n2
–
–
–
–
–
– – n X + n2X2 n1 (X2 – X1 )
and
d2 = X2 – X = X2 – 1 1
=
n1 + n2
n1 + n2
Rewriting (6·35) and substituting the values of d1 and d2 , we get
(n1 + n2) σ2 = n1 σ1 2 + n2σ2 2 + n1d1 2 + n2d2 2
–
–
n1 n2 (X1 – X2 )2
2
2
= n1 σ1 + n2σ2 +
(n1 + n2 )
(n1 + n2)2
–
–
n n (X – X2 )2
= n1 σ1 2 +n2 σ2 2 + 1 2 1
n1 +n2
–
–
n 2 σ 2 + n2σ2 2 n1 n2 (X1 – X2 )2 1/2
⇒
σ = 1 1
+
n1 + n2
(n1 + n2 )2
[
… (6·31)
]
… (6·31a)
Thus, for two groups, the formula (6·35a) can be used with convenience, since all the values are
already given.
Example 6·27. The means of two samples of sizes 50 and 100 respectively are 54·1 and 50·3 and the
standard deviations are 8 and 7. Obtain the standard deviation of the sample of size 150 obtained by
combining the two samples.
Solution. In the usual notations we are given :
–
–
n1 = 50, n2 = 100, X1 = 54·1, X2 = 50·3, σ1 = 8, σ2 = 7.
–
The mean X of the combined sample of size 150 obtained on pooling the two samples is given by :
–
–
–
n X +n X
+ 100 × 50·3 2705 + 5030 7735
X = 1 1 2 2 = 50 × 54·1
=
= 150 = 51·57.
50 + 100
150
n1 + n2
–
–
– –
d1 = X1 – X = 54·10 – 51·57 = 2·53; d2 = X2 – X = 50·30 – 51·57 = – 1·27
Hence, the variance σ2 of the combined sample of size 150 is given by :
(n1 + n2) σ2 = n1 (σ12 + d12 ) + n2 (σ22 + d22 )
⇒
150σ2 = 50[82 + (2·53)2 ] + 100 [72 + (– 1·27)2]
= 50(64 + 6·3504) + 100(49 + 1·6129)
+ 5061·29 8578·81
σ2 = 3517·52150
= 150 = 57·1921
∴
⇒
σ=√
⎯⎯⎯⎯⎯⎯
57·1921 = 7·5625.
Example 6·28. The mean weight of 150 students is 60 kgs. The mean weights of boys and girls are 70
kgs. and 55 kgs respectively , and the standard deviations are 10 kgs. and 15 kgs. respectively. Find the
number of boys and the combined standard deviation.
Solution. In the usual notations, we are given :
n = n + n =150; –x = 60 kg. ; –x = 70 kg; –x = 55 kg ; σ = 10 kg. ; σ =15 kg.
… (*)
1
2
1
2
1
where the subscripts 1 and 2 refer to boys and girls respectively. We have :
–
–
70n1 + 55n2
–x = n1 x 1 +n2 x 2
⇒
= 60
n1 + n2
150
2
[From (*)]
6·35
MEASURES OF DISPERSION
∴ 14n1 + 11 (150 – n1 ) = 60 × 30 = 1800
⇒
⇒
3n1 = 1800 – 1650 = 150
n1 =
150
= 50
3
∴
n2 = 150 – n1 = 150 – 50 = 100.
Hence, the numbr of boys is n1 = 50 and the number of girls is n2 = 100.
The combined variance (for boys and girls together) is given by :
1
σ2 =
n (σ 2 + d1 2 ) + n2 (σ2 2 + d22 )
n1 + n2 1 1
where
d = –x – –x = (70 – 60) kg. = 10kg. ;
d = –x – –x = (55– 60) kg. = – 5 kg.
[
1
]
1
2
[
2
]
1
+ 25000 700
σ2 =
50 (102 + 102) + 100 (152 + (– 5)2) = 10000150
= 3 = 233·33
150
∴
s.d. (σ) = ⎯
√⎯⎯⎯⎯
233·33 = 15·275.
Example 6·29. For a group containing 100 observations, the arithmetic mean and the standard
deviation are 8 and √
⎯⎯⎯
10·5 respectively. For 50 observations selected from these 100 observations, the
mean and standard deviation are 10 and 2 respectively. Calculate values of the mean and standard
deviation for the other half.
Solution. In the usual notations, we are given :
n = 100, –x = 8, σ = ⎯√⎯⎯⎯
10·5
⇒ σ2 = 10·5; n = 50, –x = 10, σ = 2; n = 100 – 50 = 50
⇒
1
1
1
2
We want –x2 and σ2.
–
–
–x = n1 x 1 + n2x2
n1 + n2
800 = 500 + 50x–
–
50 × 10 + 50 × x2
100
⇒
8
⇒
50x–2 = 800 – 500 = 300
⇒
–x = 300 = 6
2
50
d1 = –x1 – –x = 10 – 8 = 2 ⇒ d1 2 = 4 ;
d2 = –x2 – –x = 6 – 8 = – 2
(n1 + n2) σ2 = n1 (σ12 + d12 ) + n2 (σ22 + d22 )
⇒
100 × 10·5 = 50 (4 + 4) + 50 (σ22 + 4)
⇒
d2 2 = 4
∴
⇒
2
1050 = 400 + 50σ2 2 + 200
=
⇒
σ2 2 = 105050– 600 = 450
50 = 9
⇒
σ2 = 3,
since s.d. is always positive.
Example 6·30. Find the missing information from the following :
Group I
Group II
Group III
Combined
Number
50
?
90
200
Standard Deviation
6
7
?
7·746
Mean
113
?
115
116
(Himachal Pradesh Univ. B.Com., 1998)
Solution.
Group 1
Group 2
Group 3
Combined Group
Number
n1 = 50
n2 = ?
n3 = 90
n1 + n2 + n3 = 200
s.d.
σ1 = 6
σ2 = 7
σ3 = ?
σ = 7·746
–
–
–
–
Mean
X1 = 113
X2 = ?
X3 = 115
X = 116
–
We have three unknown values viz., n2 , σ 3 and X2 . To determine these three values we need three
equations which are given below :
6·36
BUSINESS STATISTICS
n1 + n2 + n3 = 200 …(i)
;
–
–
–
– n1 X1 + n2X2 + n3X3
X=
n1 + n2 + n3
… (ii)
and (n1 +n2 + n3 ) σ2 = n1 (σ12 + d1 2 ) + n2 (σ22 + d2 2 ) + n3 (σ32 + d3 2 )
… (iii)
From (i) we get
… (iv)
n2 = 200 – (n1 + n3 ) = 200 – (50 + 90) = 60
Using (ii) we get :
–
200 × 116 = 50 × 113 + 60 X2 + 90 × 115
∴
–
⇒
23200 = 5650 + 60X2 + 10350
–
⇒
60X2 = 23200 – (5650 + 10350)
= 23200 – 16000 = 7200
–
⇒
X2 = 7200
… (v)
60 = 120
– –
d1 = X1 – X = 113 – 116 = – 3
– –
d2 =X2 – X = 120 – 116 = 4
–
–
d3 = X3 – X = 115 – 116 = – 1
Substituting these values in (iii), we get
200 × (7·746)2 = 50(36 + 9) + 60(49 + 16) + 90(σ32 + 1)
⇒
200 × 60·000516 = 50 × 45 + 60 × 65 + 90 + 90σ3 2
⇒
12000 = 2250 + 3900 + 90 + 90σ3 2
⇒
σ3 2 = 1200090– 6240 = 5760
90 = 64
Hence, the unknown constants are :
n2 = 60,
⇒
–
X2 = 120
σ3 = 8
and
σ3 = 8.
6·11. COEFFICIENT OF VARIATION
Standard deviation is only an absolute measure of dispersion, depending upon the units of
measurement. The relative measure of dispersion based on standard deviation is called the coefficient of
standard deviation and is given by :
σ
Coefficient of Standard Deviation =
… (6·32)
X
This is a pure number independent of the units of measurement and thus, is suitable for comparing the
variability, homogeneity or uniformity of two or more distributions.
We have already discussed the relative measures of dispersion based on range, quartile deviation and
mean deviation. Since standard deviation is by far the best measure of dispersion, for comparing the
homogeneity or heterogeneity of two or more distributions, we generally compute the coefficient of
standard deviation unless asked otherwise.
100 times the coefficient of dispersion based on standard deviation is called the coefficient of variation,
abbreviated as C.V. Thus,
σ
C.V. =100 ×
… (6·33)
X
According to Professor Karl Pearson who suggested this measure, “coefficient of variation is the
percentage variation in mean, standard deviation being considered as the total variation in the mean”.
For comparing the variability of two distributions we compute the coefficient of variation for each
distribution. A distribution with smaller C.V. is said to be more homogeneous or uniform or less variable
than the other and the series with greater C.V. is said to be more heterogeneous or more variable than the
other.
Remark. Some authors define coefficient of variation as the “coefficient of standard deviation
expressed as a percentage”. For example, if mean = 15 and s.d. = 3, then
MEASURES OF DISPERSION
6·37
s.d.
= 3 = 0.20
⇒
C.V. = 20%
… (6.33a)
Mean 15
If we are given the coefficient of variation as a percentage, (like 25% or 10%), then we will use the
formula (6·33a).
C.V. =
Example 6·31. Comment on the following :
For a set of 10 observations : mean = 5, s.d. = 2 and C.V. = 60%.
6·37
MEASURES OF DISPERSION
Solution. We are given :
Using (6.37a), we get
n = 10, –x = 5 and σ = 2
σ
C.V. = – = 52 = 0·40 = 40%
x
But we are given C.V. = 60%. Hence, the given statement is wrong.
–
Example 6·32. If N = 10, X = 12, ∑X2 = 1530, find the coefficient of variation.
Solution. We have :
–
1
2
σ2 = ∑X2 – (X)2 = 1530
⇒
σ=3
10 – (12) = 153 – 144 = 9
n
[Negative sign is rejected since s.d. is always non-negative]
∴
C.V. = 100—× σ = 10012× 3 = 25
[Using (6·37)]
X
Example 6·33. The arithmetic mean of runs scored by three batsmen—Vijay, Subhash and Kumar in
the same series of 10 innings are 50, 48 and 12 respectively. The standard deviation of their runs are
respectively, 15, 12 and 2. Who is the most consistent of the three ? If one of the three is to be selected, who
will be selected ?
– – –
Solution. Let X1 , X2 , X3 be the means and σ1 , σ2 , σ3 the standard deviation of the runs scored by Vijay,
Subhash and Kumar respectively. Then we are given :
–
–
–
X1 = 50, X = 48,
X3 = 12; σ1 = 15, σ2 = 12,
σ3 = 2
C.V. of runs scored by Vijay =
100σ1
—
= 100 × 15 = 30
50
X1
100σ2
C.V. of runs scored by Subhash =
C.V. of runs scored by Kumar =
—
X2
100σ3
—
X3
= 10048× 12 = 25
= 10012× 2 = 16.67
The decision regarding the selection of player may be based on two considerations :
(i) If we want a consistent player (which is statistically sound decision), then Kumar is to be selected,
since C.V. of the runs is smallest for Kumar.
(ii) If we want to select a player whose expected score is the highest, then Vijay will be selected.
Remark. In fact, the best way will be to select a person who is consistent and also has the highest
expected score.
Example 6·34. A batsman Mr. A is more consistent in his last 10 innings as compared to another
batsman Mr. B. Therefore, Mr. A is also a higher run getter”. Comment. [Delhi Univ. B.Com. (Pass), 1999]
Solution. The consistency of a batsman is judged on the basis of coefficient of variation. The batsman
A is more consistent than batsman B if
–x
σ
σ
σA σB
σ
A
C.V. (A) < C.V. (B)
⇒
100 – A < 100 – B
⇒
<
⇒
> A
… (i)
–
–
–
σB
x
x
x
x
x
A
B
The batsman A will be higher run getter than batsman B if
–x
–x > –x
A
⇒
A
B
–x > 1
A
B
B
… (ii)
B
Since, in general, (i) does not necessarily imply (ii), the given statement is not true, in general.
In order to conclude that A is also a higher run getter, we must be given the values of –x A and –x B and
they should satisfy (ii).
6·38
BUSINESS STATISTICS
Example 6·35. Coefficien of variation of two series are 75% and 90% and their standard deviations
are 15 and 18 respectively. Find their means.
Solution. We are given : σ1 = 15 and σ2 = 18
C.V. of I series
75
= 75% = 100
Using (6·37a), we have :
C.V. =
∴
Mean of I series
=
and
Mean of II series
=
σ1
C.V. (I)
σ2
C.V.(II)
90
C.V. of II series = 90% = 100
;
σ
Mean
=
⇒
15 × 100
=
75
Mean =
σ
C.V.
20;
=
18 × 100
=
90
20
Example 6·36. “After settlement the average weekly wage in a factory had increased from Rs. 8,000 to
Rs. 12,000 and the standard deviation had increased from Rs. 100 to Rs. 150. After settlement the wage has
become higher and more uniform.” Do you agree ?
Solution. It is given that after settlement the average weekly wages of workers have gone up from
Rs. 8,000 to Rs. 12,000. This implies that the total wages received per week by all the workers together
have increased. However, we cannot conclude that the wage of each individual has increased.
Regarding uniformity of the wages, we have to calculate the coefficient of variation of the wages of
workers before the settlement and after the settlement.
× 100
C.V. of wages before the settlement = 100
8‚000 = 1.25
× 150
C.V. of wages after the settlement = 100
12‚000 = 1.25
Since the coefficient of variation of wages before the settlement and after the settlement is same, there
is no change in the variability of distribution of wages after the settlement. Hence, it is wrong to say that the
wages have become more uniform (less variable) after the settlement.
Example 6·37. A study of B.A. (Hons.) Economics examination results of 1000 students in 1990 gave
the mean grade as 78 and the standard deviation as 8·0. A similar study in 1995 revealed the mean grade
of the group as 80 and the standard deviation as 7·6. In which year was there the greater
(i) absolute dispersion,
(ii) relative dispersion ?
What can we say about the average performance of the students over time ?
[Delhi Univ. B.A. (Econ. Hons.), 1999]
Solution. In the usual notations, we are given :
–
–
Year 1990 : X1 = 78; σ1 = 8·0;
Year 1995 : X2 = 80, σ2 = 7·6
(i) We know that the best absolute measure of dispersion is the standard deviation.
Since σ1 > σ2 , in 1990 there was greater absolute dispersion.
(ii) The relative measure of dispersion is given by the coefficient of variation.
100 σ
100 σ
× 7·6
C.V. (1990) = – 1 = 10078× 8.0 = 10.26
;
C.V. (1995) = – 2 = 10080
= 9·50
X1
X2
Since C.V. (1990) > C.V. (1999), the relative dispersion is greater in 1990.
– –
We observe that X2 > X1 and C.V. (1995) < C.V. (1990). Hence, it can be said that over the time from
1990 to 1995, the average performance of the students has increased (improved) and they have become
more consistent (less variable).
Example 6.38. Explain the difference between absolute and relative dispersion. If 20 is subtracted
from every observation in a data set, then the coefficient of variation of the resulting data set is 20%. If 40
is added to every observation of the same data set, then the coefficient of variation of the resulting set of
data is 10%. Find the mean and standard deviation of the original set of data.
[Delhi Univ. B.Com. (Hons.), 2004]
6·39
MEASURES OF DISPERSION
—
Solution. Let X be the mean and σX = σ, be the standard deviation of the data observations
Let U = X – 20 ⇒
—
—
and V = X + 40 ⇒
—
—
and
U = X – 20 and σU = σX = σ
} [Q
V = X + 40 and σV = σX = σ
100 σ
100 σ
C.V. (U) = — U = —
= 20 (Given)
U
X – 20
100 σ
100 σ
C.V. (V) = — V = —
= 10 (Given)
V
X + 40
s.d. is independent of
change of origin
]
…(i)
…(ii)
—
—
X + 40
=2
⇒
X = 800
—
X – 20
100σ
Substituting in (i), we get
= 20
⇒
σ = 12
60
Example 6·39. An analysis of the monthly wages paid to workers in two firms A and B, belonging to
the same industry, gives the following results :
Firm A
Firm B
Number of wages earners
550
650
Average monthly wages (in ’00 Rs.)
50
45
Standard deviation of the distribution of wages (in ’00 Rs.)
⎯⎯90
√
⎯⎯⎯
√
120
Dividing (i) by (ii) we get
Answer the following questions with proper justifications :
(a) Which firm A or B pays larger amount as monthly wages ?
(b) In which firm A or B, is there greater variability in individual wages ?
(c) What are the measures of (i) average monthly wages and (ii) standard deviation in the
distribution of individual wages of all workers in the two firms taken together ?
– –
Solution. Let n1 , n 2 denote the sizes ;X1 , X2 the means and σ1 , σ2 the standard deviations of the
monthly wages (in Rs.) of the workers in the firms A and B respectively. Then we are given :
–
n1 = 550,
X1 = 50,
σ1 = ⎯√⎯90
⇒
σ1 2 = 90
–
n2 = 650,
X2 = 45,
σ2 = ⎯√⎯⎯
120
⇒
σ2 2 = 120
(a) We know that
Total monthly wages paid by the firm
Average monthly wages =
No. of workers in the firm
⇒ Total monthly wages paid by the firm = (No. of workers in the firm) × (Average monthly wages)
–
∴
Total monthly wages paid by firm A = n1 X1 = Rs. 550 × 50 = Rs. 27,500 hundred
–
Total monthly wages paid by firm B = n2 X2 = Rs. 650 × 45 = Rs. 29,250 hundred
Hence the firm B pays out larger amount as monthly wages, the excess of the monthly wages paid over
firm A being
Rs. (29,250 – 27,500) hundred = Rs. 1,750 hundred
(b) In order to find out which firm has more variation in individual wages, we have to compute the
coefficient of variation (C.V.) of the distribution of monthly wages for each of the two firms A and B.
C.V. for firm A =
C.V. for firm B =
100 σ1
—
X1
100 σ2
—
X2
= 10050√⎯⎯90 = 2 × 9·487 = 18·974
⎯⎯⎯
√
120
= 10045
=
20 × 10·954 219·080
=
9
9
= 24·34
6·40
BUSINESS STATISTICS
Since C.V. for firm B is greater than the C.V. for firm A, firm B has greater variability in individual
wages.
–
(c) (i) The average monthly wage, say,X, of all the workers in the two firms A and B taken together is
given by :
—
—
– n1 X 1 + n2X 2
X = n +n
500 × 50 + 650 × 45
hundred
550 + 650
27‚500 + 29‚250
Rs.
hundred = Rs. 56‚750
1‚200
1‚200 hundred
1
=
= Rs.
2
= Rs. 47·29 hundred = Rs. 4,729.
(ii) The variance σ2 of the distribution of monthly wages of all the workers in the two firms A and B
taken together is given by :
(n1 + n2) σ2 = n1 (σ12 + d12 ) + n2 (σ22 + d22 )
… (*)
– –
d1 =X1 – X = 50 – 47·29 = 2·71
⇒
d1 2 = 7·344
–
–
and
d2 = X2 – X = 45 – 47·29 = – 2·29
⇒
d2 2 = 5·244
Substituting in (*), we get
1200σ2 = 550 (90 + 7·344) + 650 (120 + 5·244) = 550 × 97·344 + 650 × 125·244
= 53539·2 + 81408·6 = 134947·8
134947·8
2
⇒
σ = 1200 = 112·4565 Rs.2 ⇒ σ = Rs. ⎯√⎯⎯⎯⎯⎯
112·4565 hundred = Rs. 10·60 hundred = Rs. 1060
Example 6·40. From the prices X and Y of shares A and B respectively given below, state which share
is more stable in value.
Price of Share A (X) :
55
54
52
53
56
58
52
50
51
49
Price of Share B (Y) :
108
107
105
105
106
107
104
103
104
101
Solution.
COMPUTATION OF MEAN AND S.D. OF PRICES OF SHARES A AND B
SHARE A
SHARE B
—
—
—
—
X
X – X = X – 53
(X – X ) 2
Y
Y – Y = Y – 105
(Y – Y ) 2
55
2
4
108
3
9
54
1
1
107
2
4
52
–1
1
105
0
0
53
0
0
105
0
0
56
3
9
106
1
1
58
5
25
107
2
4
52
–1
1
104
–1
1
50
–3
9
103
–2
4
51
–2
4
104
–1
1
49
∑X = 530
–4
16
101
∑Y = 1050
–4
0
—
∑(X – X ) = 0
– ∑X 530
X = n = 10 = 53
—
∑(X – X ) = 70
2
16
—
∑(Y – Y ) 2 = 40
,
–
σx2 = 1n ∑(X – X) 2 = 70
10 = 7
⇒
σx = ⎯
√ 7 = 2·646
– ∑Y 1050
Y = n = 10 = 105 ,
–
40
σy2 = 1n ∑(Y – Y ) 2 = 10
=4
⇒
σy = ⎯
√4 = 2
C.V. (X) =
100 × σx
—
X
=
100 × 2.646
=
53
4·99
;
C.V. (Y) =
100 × σy
—
Y
=
100 × 2
105 =
Since C.V. (Y) is less than C.V. (X), the share B is more stable in value.
1·90
6·41
MEASURES OF DISPERSION
Example 6·41. Goals scored by two teams
A and B in a football season were as shown in
adjoinging table.
By calculating the coefficient of variation
in each case, find which team may be
considered more consistent.
Number of goals
scored in a match
0
1
2
3
4
Number of matches
A team
B team
27
17
9
9
8
6
5
5
4
3
Solution.
CALCUALTIONS FOR C.V. FOR TEAMS A AND B
TEAM A
TEAM B
X2
No. of goals scored in
a match (X)
No. of
matches (f1)
f 1X
f1
No. of
matches (f2)
f 2X
0
1
2
3
4
27
9
8
5
4
0
9
16
15
16
0
9
32
45
64
17
9
6
5
3
0
9
12
15
12
Total
∑ f 1 = 53
∑ f 1 X = 56
∑ f 2 = 40
∑ f 2X = 48
–
∑f X 56
XA = 1 = 53
= 1·06
∑f1
∑f X2 –
2
σA2 = 1 – (XA)2 = 150
53 – (1·06)
∑f1
= 2·83 – 1·1236 = 1·7064
σA = √⎯⎯⎯⎯⎯
1·7064 = 1·3063
100 × σA
∴ C.V. for team A = –
XA
∑ f 1X2 = 150
f 2X2
0
9
24
45
48
∑ f 2X2 = 126
–
∑f X 48
XB = 2 = 40
= 1·2
∑f2
∑f X2 –
2
σB2 = 2 – (XB)2 = 126
40 – (1·2)
∑f2
= 3·15 – 1·44 = 1·71
⇒
σB = √⎯⎯⎯
1·71 = 1·308
100 × σB
∴ C.V. for team B =
–
XB
× 1·3063
= 100 1·06
= 123·24
= 100 ×1·21·308 = 109·0
Since C.V. for team B is less then C.V. for team A, team B may be considered to be more consistent.
6·12. RELATIONS BETWEEN VARIOUS MEASURES OF DISPERSION
For a Normal distribution (c.f. Chapter 14 on theoretical distributions) we have the following relations
between the different measures of dispersion :
(i) Mean ± Q.D. covers 50% of the observations of the distribution.
(ii) Mean ± M.D. covers 57·5% of the observations.
(iii) Mean ± σ includes 68·27% of the observations.
(iv) Mean ± 2σ includes 95·45% of the observations.
(v) Mean ± 3σ includes 99·73% of the observations.
2
(vi) Q.D. = 0·6745σ ≈ 3 σ (approximately).
2
π
(vii) M.D. =
(viii) Q.D. = 0·8459 M.D.
5
6
4
. σ = 0·7979 ≈ 5 σ (approximately).
… (6·34)
… (6·35)
[From (vi) and (vii), on dividing and transposing]
⇒ Q.D. = M.D. (approximately)
… (6·36)
6·42
BUSINESS STATISTICS
Combining the results (vi), (vii), and (viii) we get approximately :
3 Q.D. = 2 S.D.
;
5 M.D. = 4 S.D.
;
6 Q.D. = 5 M.D.
⇒
4 S.D. = 5 M.D. = 6 Q.D.
Thus we see that standard deviation ensures the highest degree of reliability and Q.D. the lowest.
(ix) We have :
Q.D. : M.D. : S.D. : : 23 σ : 45 σ :
σ
⇒
Q.D. : M.D. : S.D. : : 10 : 12 :
15
…(6·37)
(x) Range = 6 S.D. = 6σ
…(6·38)
Remarks 1. Rigorously speaking, the above results for various measures of dispersion hold for Normal
distribution discussed in Chapter 14 on theoretical distributions. However, these results are approximately
true even for symmetrical distributions or moderately asymmetrical (skewed) distributions.
2. In the above results we have expressed various measures of dispersion in terms of standard
deviation. We give below the relations expressing standard deviation in terms of other measures of
dispersion.
S.D. = 1·2533 M.D. –~ 45 M.D.
}
S.D. = 1·4826 Q.D. ~– 32 Q.D.
S.D. = 61 Range
Also we have : M.D. = 1·1830 Q.D. ~– 65 Q.D.
…(6·39)
…(6·40)
EXERCISE 6·4
1. (a) What do you understand by absolute and relative measure of dispersion ? Explain advantages of the relative
measures over the absolute measures of dispersion.
(b) What do you understand by coefficient of variation ? What purpose does it serve ?
2. Prove that the coefficient of variation of the first n natural numbers is :
(n – 1)
.
3 (n + 1)
[Delhi Univ. B.A. Econ. (Hons.), 2006, 2003]
3. The arithmetic means of runs secured by the three batsmen, X, Y and Z in a series of 10 innings are 50, 48 and
12 respectively. The standard deviations of their runs are 15, 12 and 2 respectively. Who is the most consistent of the
three ?
Ans. C.V. (X) = 30 ; C.V. (Y) = 25 ; C.V. (Z) = 16·67. Batsman Z is the most consistent.
4. Two samples A and B have the same standard deviations, but the mean of A is greater than that of B. The
coefficient of variation of A is
(i) greater than that of B.
(ii) less than that of B.
(iii) equal to that of B.
(iv) None of these.
Ans. (i)
5. (a) The coefficient of variation of a distribution is 60% and its standard deviation is 12. Find out its mean.
(b) Find the coefficient of variation if variance is 16, number of items is 20 and sum of the items is 160.
[Bangalore Univ. B.Com., 1998]
Ans. (a)Mean = 20 ; (b) C.V. = 50%.
6. (a)Coefficients of variation of two series are 60% and 80%. Their standard deviations are 20 and 16
respectively. What are their arithmetic means ?
(b) Coefficients of variation of two series are 60% and 80%. Their standard deviations are 24 and 20 respectively.
What are their arithmetic means ?
[Delhi Univ. B.Com. (Pass), 1997]
Ans. (a) 33·3, 20; (b) 40, 25.
7. Comment on the statement : “After settlement the average weekly wage in a factory had increased from Rs. 800
to Rs. 1200 and the standard deviation had increased from 200 to 250. After settlement, the wages have become higher
and more uniform.
Ans. Yes.
6·43
MEASURES OF DISPERSION
8. Weekly average wages of workers in a factory increase from Rs. 800 to Rs. 1200 and standard deviation
increases from Rs. 100 to Rs. 500. Have the wages become less uniform now ?
Ans. C.V.(Initial wages) = 12·5 ; C.V. (Revised wages) = 41·67. Yes, the revised wages are more variable.
9. A study of examination results of a batch of students showed the average marks secured as 50 with a standard
deviation of 2 in the first year of their studies. The same batch showed an average of 60 marks with an increased
standard deviation of 3, after five years of studies. Can you say that the batch as a whole showed improved
performance ?
Ans. Improved performance (better average) and more consistent.
10. The means and standard deviations of two brands of light bulbs are given below :
Brand 1
800 hours
100 hours
Mean
S.D.
Brand 2
770 hours
60 hours
Calculate a measure of relative dispersions for the two brands and interpret the results.
[Delhi Univ. B.Com. (Hons.), 2000]
Ans. C.V. (I) = 12·5; C.V. (II) = 7·79 ; Brand II is more uniform.
11. The following is the record number of bricks laid each day for 10 days by two bricklayers A and B. Calculate
the coefficient of variation in each case and discuss the relative consistency of the two bricklayers.
A
700
675
725
625
650
700
650
700
600
650
B
550
600
575
550
650
600
550
525
625
600
If each of the values in respect of worker A is decreased by 10 and each of the values for worker B is increased by
50, how will it affect the results obtained earlier ?
[Delhi Univ. B.Com. (Hons), 2007]
Ans.
—
X A = 667.5
C.V. (A) =
—
σA = 37.165 ;
37.165
× 100 = 5.57
667.5
C.V. (A) < C.V. (B)
X B = 582.5,
;
⇒
C.V. (B) =
σB = 37.165
37.165
× 100 = 6.37
582.5
Brick layer A is more consistent.
Since S.D. is independent of change of origin, we have :
—
New Mean : X A ′ = 667.5 – 10 = 657.5
σA′ = σA = 37.165
;
;
—
X B ′ = 582.50 + 50 = 632.50
σB′ = σB = 37.165
[New C.V. (A) = 5.65] < [ New C.V. (B) = 5.88]. Result is not affected.
12. The number of employees, average wages per employee and variance of the wages per employee for two
factories are given below :
Factory A
Factory B
Number of Employees
100
200
Average Wage per Employee (Rs.)
120
200
16
25
Variance of the Wages per Employee (Rs.)
In which factory is there greater variation in the distribution of wages per employee ?
[C.A. (Foundation), May 2000]
Ans. C.V.(A) = 3·3; C.V. (B) = 2·5. There is greater variability in Factory A.
13. The number of employees, wages per employee and the variance of the wages per employee for two factories
are given below :
Factory A
Factory B
No. of employees
50
100
Average wages per employee per month (in Rs.)
120
85
Variance of the wages per employee per month (in Rs.)
9
16
6·44
BUSINESS STATISTICS
In which factory is there greater variation in the distribution of wages per employee ?
Ans. In factory B ; C.V. (A) = 2·5, C.V. (B) = 4·7.
14. Two workers on the same job show the following results over a long period of time :
Worker A
Mean time of completing the job (minutes)
30
Standard deviation (minutes)
6
Worker B
25
4
(i) Which worker appears to be more consistent in the time he requires to complete the job ?
(ii) Which worker appears to be faster in completing the job ? Explain.
—
—
Ans. (i) B, (ii) B (·.· X B < X A ).
15. The mean and standard deviation of 200 items are found to be 60 and 20 respectively. At the time of
calculations, two items were wrongly taken as 3 and 67 instead of 13 and 17. Find the correct mean and standard
deviation. What is the correct coefficient of variation ?
Ans. Corrected mean = 59·8, s.d. = 20·09; C.V. = 33·60.
16. The mean and standard deviation of a series of 100 items were found to be 60 and 10 respectively. While
calculating, two items were wrongly taken as 5 and 45 instead of 30 and 20. Calculate corrected variance and
corrected coefficient of variation.
[Delhi Univ. B.Com. (Hons.), 2009]
Ans. Corrected (σ2) = 92.50
;
Corrected C.V. = 16.03
17. For the following distribution of marks obtained, find the arithmetic mean, the standard deviation and the
coefficient of variation.
Marks obtained :
0—5
5—10
10—15
15—20
20—25
25—30
30—35
35—40
No. of students :
2
5
7
13
21
16
8
3
Ans. A.M. = 21·9; σ = 7·9931; C.V. = 36·5.
18. Data on the annual earnings of professors and physicians in a certain town yield the following results :
–
–
Professors : x 1 = Rs.16,000,
σ1 = Rs. 2,000
Physicians : x 2 = Rs.23,000,
σ2 = Rs. 4,000
Are the professors’ earnings more or less variable than the physicians’ ?
[Delhi Univ. B.A. (Econ. Hons.), 1990]
Ans. C.V. (Professors) = 12·50, C.V. (Physicians) =17·39; Professors’ earnings are less variable.
19. (a) Verify the correctness of the following statement : “A batsman scored at an average of 60 runs an inning
against Pakistan. The standard deviation of the runs scored by him was 12. A year later against Australia, his average
came down to 50 runs an inning and the standard deviation of the runs scored fell down to 9. Therefore, it is correct to
say that his performance was worse against Australia and that there was lesser consistency in his batting against
Australia”.
[Delhi Univ. B.Com. (Hons.), 1986]
Ans. C.V. (Australia) = 18, C.V. (Pakistan) = 20. Greater consistency against Australia.
—
—
X (Against Pakistan) > X (Against Australia); better performance against Pakistan than against Australia.
(b) The following is the record of goals scored by team A in the football season :
No. of goals scored by team A in a match
:
0
1
2
3
4
Number of matches
:
1
9
7
5
3
For team B the average number of goals scored per match was 2·5 with a standard deviation of 1·25 goals.
Find which team may be considered more consistent.
Ans. C.V. (A) = 54·77, C.V. (B) = 50; B is more consistent.
20. During the 10 weeks of a session, the marks obtained by two candidates, Ramesh and Suresh, taking the
Computer Programme course are given below :
Ramesh :
58
59
60
54
65
66
52
75
69
52
Suresh :
87
89
78
71
73
84
65
66
56
46
(i) Who is the better scorer — Ramesh or Suresh ?
(ii) Who is more consistent ?
[Delhi Univ. B.Com. (Pass), 1998]
–
–
Ans. Ramesh : x1 = 61,
σ1 = 7·25,
C.V. = 11·89 ; Suresh : x 2 = 71·5,
σ2 = 13·08,
C.V. = 18·29
– –
(i) Suresh is better scorer (·.· x 2 > x 1). ; (ii) Ramesh is more consistent.
6·45
MEASURES OF DISPERSION
21. Complete the table showing the frequencies with which words of different number of letters occur in the
extract reproduced below (omitting punctuation marks) treating as the variable the number of letters in each word, and
obtain the coefficient of variation of the distribution :
“Her eyes were blue; blue as autumn distance—blue as the blue we see between the retreating mouldings of hills
and woody slopes on a sunny September morning : a misty and shady blue, that had no beginning or surface, and was
looked into rather than at”.
—
Ans. X = 4·35; σ = 2·23; C.V. = 51·04.
22. Compile a table, showing the frequencies with which words of different number of letters occur in the extract
reproduced below (omitting punctuation marks) treating as the variable the number of letters in each word, and obtain
the mean, median and the coefficient of variation of the distribution :
“Success in the examination confers no absolute right to appointment unless Government is satisfied, after such
enquiry as may be considered necessary, that the candidate is suitable in all respects for appointment to the public
service”.
Ans. Mean = 5·5; Median = 5; s.d. = 3·12; C.V. = 56·7.
23. No of goals scored in a match :
0
1
2
3
4
Team A :
27
9
8
5
1
No. of Matches
Team B :
1
5
8
9
27
]
Ans. C.V. (A) = 127·84;
C.V. (B) = 36·06;
Team B is more consistent.
Life
24. Lives of two models of refrigerators in a recent survey are
shown in adjoinging table.
What is the average life of each model of these refrigerators ?
Which model has greater uniformity ?
—
—
Ans. X a = 5·12 years ; C.V. (A) = 54·88; X b = 6·16 years.
C.V. (B) = 36.2; Model B has greater uniformity.
No. of refrigerators
(No. of years)
0—2
2—4
4—6
6—8
8—10
10—12
Model A
5
16
13
7
5
4
Model B
2
7
12
19
9
1
25. A purchasing agent obtained samples of incandescent lamps from two suppliers. He had the samples tested in
his own labouratory for the length of life with the following results :
Length of life in hours
Samples from
Company A
Company B
700 and under 900
10
3
900 and under 1,100
16
42
1,100 and under 1,300
26
12
1,300 and under 1,500
8
3
Which company’s lamps are more uniform ?
Ans. C.V. (A) = 16·7, C.V. (B) = 11·9. Lamps of company B are more uniform.
26. Two brands of tyres are tested with the following results :
Life (in ’000 miles)
(a) Which brand of tyres have greater average life ?
(b) Compare the variability and state which brand of tyres would
you use on your fleet of trucks ?
20—25
25—30
30—35
35—40
40—45
No. of tyres of brand
X
1
22
64
10
3
Y
0
24
76
0
0
[Bangalore Univ. B.Com., 1997]
—
Ans. X = 32·1 thousand miles,
—
σx = 3·137 thousand miles,
C.V. (X) = 9·77;
Y = 31·3 thousand miles, σy = 0·912 thousand miles, C.V. (Y) = 2·914.
(a) Brand X ; (b) Brand X tyres are more variable; Brand Y.
6·46
BUSINESS STATISTICS
27. The mean and standard deviation of the marks obtained by two groups of students, consisting of 50 each, are
given below. Calculate the mean and standard deviation of the marks obtained by all the 100 students :
Group
Mean
Standard Deviation
1
60
8
2
55
7
[C.A. (Foundation), Nov. 1999]
–
Ans. x 12 = 57·5, σ12 = 7·92.
28. For a group of 50 male workers, the mean and standard deviation of their weekly wages are Rs. 63 and Rs. 9
respectively. For a group of 40 female workers these are Rs. 54 and Rs. 6 respectively. Find the standard deviation for
the combined group of 90 workers.
Ans. σ = 9.
29. The first of the two samples has 100 items with mean 15 and standard deviation 3. If the whole group has 250
√⎯⎯⎯
13·44, find mean and the standard deviation of the second group.
items with mean 15·6 and standard deviation ⎯
—
Ans. X 2 = 16; σ 2 = 4.
30. For two groups of observations the following results were available :
Group I :
∑(X – 5) = 8;
∑(X – 5)2 = 40;
N1 = 20
Group II :
∑(X – 8) = – 10;
∑(X – 8)2 = 70;
N2 = 25
Find the mean and the standard deviation of the 45 observations obtained by combining the two groups.
[Delhi Univ. B.Com. (Hons.), 2006]
Hint. Expand ∑(X – 5), ∑(X – 5)2, ∑(X – 8) and ∑(X – 8)2, and find ∑X and ∑X 2 for each group.
For combined group :
—
∑X = 108 + 190 = 298,
∑X2 = 620 + 1510 = 2130,
N= 45.
Ans. X 12 = 6·622, σ12 = 1·865.
31. A company has three establishments E1, E 2, E 3 in three cities. Analysis of the monthly salaries paid to the
employees in the three establishments is given bellow :
E1
E2
E3
Number of employees
20
25
40
Average monthly salary (Rs.)
305
300
340
Standard deviation of monthly salary (Rs.)
50
40
45
Find the average and the standard deviation of the monthly salaries of all the 85 employees in the company.
Ans. Mean = Rs. 320; s.d. = Rs. 48·69.
32. Calculate the missing information from the following data.
Variable A
Variable B
Total Number
175
?
Mean
220
240
Standard Deviation
?
6·3
Variable C
225
?
5·9
Combined
500
235
5·4
—
Ans. nB = 100; X C = 244·4, σA = 18·36.
33. For a group of 30 male workers, the mean and standard deviation of weekly overtime work (No. of hours) are
10 and 4 respectively; for 20 female workers the mean and standard deviation are 5 and 3 respectively. (i) Calculate the
mean for the two groups taken together. (ii) Is the overtime work more variable for the male group than for the female
group ? Explain.
[Delhi Univ. B.A. (Econ.) Hons., 1991]
Ans. (i) 8 hours.
(ii) C.V. males (= 40) < C.V. females (= 60) ⇒ Overtime work for males is less variable.
34. An analysis of the monthly wages paid to workers in two firms A and B, belonging to the same industry, gives
the following results :
Firm A
Firm B
Number of wage-earners
586
648
Average monthly wage
Rs. 52·5
Rs. 47·5
6·47
MEASURES OF DISPERSION
Variance of the distribution of wage
100
121
(a) Which firm, A or B, pays out the larger amount as monthly wages ?
(b) In which firm, A or B, is there greater variability in individual wages ?
(c) What are the measures of (i) average monthly wage, and (ii) the variability in individual wages, of all the
workers in the two firms, A and B, taken together.
—
Ans. (a) B; (b) In Firm B; (c) X = Rs. 49·87; σ = Rs. 10·83.
35. For two firms A and B, the following details are available :
Number of employees
Average salary (Rs.)
Standard deviation of salary (Rs.)
:
:
:
A
100
1,600
16
B
200
1,800
18
(i) Which firm pays large package of salary ?
(ii) Which firm shows greater variability in the distribution of salary.
(iii) Compute the combined average salary and combined variance of both the firms.
[Delhi Univ. B.Com. (Pass), 2000]
Ans. (i) B; (ii) C.V. (A) = 1, C.V. (B) = 1. Both firms show equal variability.
–
(iii) x 12 = Rs. 1,733·33 and σ212 = Rs2. 9190·22
36. If the mean deviation of a moderately skewed distribution is 7·2 unit, find the standard deviation as well as
quartile deviation.
5
5
Ans. S.D. ~
– M.D. = 9; Q.D. –~ 6 M.D. = 6·0.
4
37. For a series, the value of Mean Deviation is 15. Find the most likely value of its quartile deviation.
[Delhi Univ. B.Com. (Pass), 2002]
5
Ans. Q.D. = M.D. = 12·5.
6
6·13. LORENZ CURVE
Lorenz curve is a graphic method of studying the dispersion in a distribution. It was first used by Max
O. Lorenz, an economic statistician for the measurement of economic inequalities such as in the distribution
of income and wealth between different countries or between different periods of time. But today, Lorenz
curve is also used in business to study the disparities of the distribution of wages, profits, turnover,
production, population, etc.
A very distinctive feature of the Lorenz curve consists in dealing with the cumulative values of the
variable and the cumulative frequencies rather than its absolute values and the given frequencies. The
technique of drawing the curve is fairly simple and consists of the following steps :
(i) The size of the item (variable value) and the frequencies are both cumulated. Taking grand total for
each as 100, express these cumulated totals for the variable and the frequencies as percentages of their
corresponding grand totals.
(ii) Now take coordinate axes, X-axis representing the percentages of the cumulated frequencies (x) and
Y-axis representing the percentages of the cumulated values of the variable (y). Both x and y take the values
from 0 to 100 as shown in the Fig. 6·1.
(iii) Draw the diagonal line y = x joining the origin O (0, 0) with the point P(100, 100) as shown in the
diagram. The line OP will make an angle of 45° with the X-axis and is called the line of equal distribution.
(iv) Plot the percentages of the cumulated values of the variable (y) against the percentages of the
corresponding cumulated frequencies (x) for the given distribution and join these points with a smooth freehand curve. Obviously, for any given distribution this curve will never cross the line of equal distribution
6·48
BUSINESS STATISTICS
C
B
Income
OP. It will always lie below OP unless the distribution is uniform (equal) in which case it will coincide
with OP.
Thus when the distribution of items is not proportionately equal, the variability (dispersion) is
indicated and the curve is farther from the line of equal distribution OP. The greater the variability, the
greater is the distance of the curve from OP.
Let us consider the Lorenz curve
LORENZ CURVE
diagram (Fig. 6.1), for the distribution of
Y
P (100,100)
income, (say).
100
•
In the diagram, OP is the line of equal
distribution of income. If the plotted
cumulative percentages lie on this line,
n
io
there is no variability in the distribution
ut
b
i
of income of persons. The points lying
str
di
l
on the curve OAP indicate a less degree of
a
qu
A
variability as compared to the points lying
fe
o
e
n
on the curve OBP. Variability is still
Li
greater, when the points lie on the curve
OCP. Thus a measure of variability of the
distribution is provided by the distance of
the curve of the cumulated percentages of
X
O
the given distribution from the line of
100
No. of persons
equal distribution.
Fig. 6·1.
Remarks 1. An obvious disadvantage of the Lorenz curve is that it gives us only a relative idea of the
dispersion as compared with the line of equal distribution. It does not provide us any numerical value of the
variability for the given distribution. Accordingly it should be used together with some numerical measure
of dispersion. However, this should not undermine the utility of Lorenz curve in studying the variability of
the distributions particularly relating to income, wealth, wages, profits, lands, and capitals, etc.
2. From the Lorenz curve we can immediately find out as to what percentage of persons (frequencies)
correspond to a given percentage of the item (variable value).
Example 6·42. From the following table giving data regarding income of workers in a factory, draw a
graph (Lorenz curve) to study the inequality of income :
Income (in Rs.)
No. of workers in the factory
Below 500
6,000
500—1,000
4,250
1,000—2,000
3,600
2,000—3,000
1,500
3,000—4,000
650
Solution.
CALCULATIONS FOR LORENZ CURVE
Income (in Rs.)
Mid-value
Cumulative
income
Percentage of
cumulative
income
No. of
Workers (f)
Cumulative
frequency
Percentage of
cumulative
frequency
(1)
(2)
(3)
(4)
(5)
(6)
(7)
0—500
500—1000
1000—2000
2000—3000
250
750
1500
2500
250
1,000
2,500
5,000
2·94
11.76
29.41
58.82
6,000
4,250
3,600
1,500
6,000
10,250
13,850
15,350
37·5
64.1
86.6
95.9
6·49
MEASURES OF DISPERSION
3000—4000
3500
Total
8500
8,500
100.00
650
16,000
100.0
16,000
The Lorenz curve (Fig. 6·2) prominently exhibits the inequality of the distribution of income among
the factory workers.
LORENZ CURVE
• (100, 100)
O
N
90 –
U
TI
80 –
D
IS
TR
IB
70 –
•
EQ
U
A
L
60 –
50 –
O
F
40 –
N
E
30 –
•
LI
20 –
–
–
–
–
–
–
–
–
O
•
–
•
10 –
–
PERCENTAGE OF INCOME →
100 –
10 20 30 40 50 60 70 80 90 100
PERCENTAGE OF PERSONS →
Fig. 6·2.
Remark. From the Lorenz curve we observe that 70% of the persons get only 15% of income and 90%
of the persons get only 35% of the income.
EXERCISE 6·5
1. What is Lorenz curve ? How do you construct it ? What is its use?
2. From the following table giving data regarding income of employees in two factories, draw a graph (Lorenz
Curve) to show which factory has greater inequalities of income :
Income (’00 Rs.) :
Below 200
200—500
500—1,000
1,000—2,000
2,000—3,000
Factory A
:
7,000
1,000
1,200
800
500
Factory B
:
800
1,200
1,500
400
200
3. Industrial relations have been deteriorating in
the Wessex factory of JK Limited and personnel
management has established that a contributory factor
is the inequalities of earnings of operators paid on the
basis of an incentive scheme.
Operators work an eight-hour day and bonus is
paid progressively after measured work equivalent to
360 standard minutes has been produced.
Table A, on right side, shows the position for the
month of June 1975 in respect of operator production.
Improvements are made in working conditions in
the areas where poor performances were recorded and
subsequently in October 1975 the results in Table B,
were measured.
Standard minutes
Table A (June
Table B (October
produced per
1975) Number of 1975) Number of
operator per day
operators
operators
4
10
300
11
32
320
11
20
340
9
18
360
10
2
380
12
5
400
12
5
420
11
5
440
10
3
460
5
—
480
5
—
500
(a) Present the data in Tables A and B in the form of a Lorenz curve.
(b) Comment on the results.
Ans. Group A represents greater inequality of distribution than group B.
4. The frequency distribution of marks obtained in Mathematics (M) and English (E) are as follows :
Mid-value of marks
:
5
15
25
35
45
55
65
75
85
95
6·50
BUSINESS STATISTICS
No. of students (M) :
10
12
13
14
22
27
20
12
11
9
No. of students (E)
:
1
2
26
50
59
40
10
8
3
1
Analyse the data by drawing the Lorenz curves on the same diagram and describe the main features you observe.
5. Draw Lorenz Curve for the comparison of profits of two groups of companies, A and B, in business. What is
your conclusion ?
Total Amount of profits earned by
companies
Number of Companies in
Group A
Group B
6
11
13
14
15
17
10
14
1
19
26
14
14
13
6
7
600
2,500
6,000
8,400
10,500
15,000
17,000
40,000
Ans. Lorenz curve for the group B is farthest from the line of equal distribution. Hence, group B represents greater
inequality of profits than group A.
6. (a)Write an explanatory note on Lorenz curve.
(b) The following table gives the population and earnings of residents in towns A and B. Represent the data
graphically so as to bring out the inequality of the distribution of the earnings of residents.
Town A
Town B
No. of persons :
50
50
50
50
50
50
50
50
50
50
Earnings (Rs. daily)
35
50
75
115
160
180
225
300
425
925
No. of persons :
100
140
60
50
200
90
60
40
160
100
Earnings (Rs. daily)
160
320
120
280
400
400
280
920
240
960
Ans. Inequality of incomes is more prominent in Town A.
EXERCISE 6·6
(Objective Type Questions)
I. Match the correct parts to make a valid statement :
(a) Algebraic sum of deviations from mean
(b) Coefficient of Mean Deviation
(c) Variance
(d) Quartile Deviation
(e) Coefficient of Variation
(f) Sum of absolute deviations from median
(i) Q3 – Q1
100 σ
Mean
(iii) Zero
—
1
∑f|X–X|
(iv)
N
—
1
(v)
∑ f (X – X )2
N
M.D. about Mean
(vi)
Mean
(vii) (Q3 – Q1)/2
(viii) Minimum
(d) — (vii);
(h) — (iv).
(ii)
(g) Interquartile range
(h) Mean deviation
Ans. (a) — (iii);
(b) — (vi);
(c) — (v);
(e) — (ii);
(f) — (viii);
(g) — (i);
II. In the following questions, tick the correct answer :
(i) Algebraic sum of deviations from mean is :
(a) Positive,
(b) Negative,
(c) Zero,
(d) Different for each case.
(ii) Sum of squares of deviations is minimum when taken from :
6·51
MEASURES OF DISPERSION
(a) Mean,
(b) Median,
(c) Mode,
(d) None of these.
(iii) Sum of absolute deviations is minimum when measured from :
(a) Mean,
(b) Median,
(c) Mode,
(d) None of these.
(iv) For a discrete frequency distribution :
(a) S.D. ≤ M.D., (b) S.D. ≥ M.D., (c) S.D. > M.D.,
(d) S.D. < M.D.
(e) None of these, where M.D. = Mean Deviation from mean.
(v) The range of a given distribution is
(a) greater than s.d.,
(b) Less than s.d., (c) Equal to s.d., (d) None of these.
(vi) The measure of dispersion independent of frequencies of the given distribution is
(a) Range,
(b) s.d.,
(c) M.D.,
(d) Q.D.
(vii) In case of open end classes, an appropriate measure of dispersion to be used is
(a) Range,
(b) Q.D.,
(c) M.D.,
(d) s.d.
(viii) Measure of dispersion which is affected most by extreme observations is.
(a) Range,
(b) Q.D.,
(c) M.D.,
(d) s.d.
(ix) Mean deviation from median (Md) is given by
(a)
∑ | X – Md |
,
n
(b)
∑ | X – Md |
,
(c)
∑ | X – Md |2
,
n
⎯n
√
(x) Quartile deviation is given by :
Q 3 – Q2
Q2 – Q1
(a) Q3 – Q1,
(b)
,
(c)
,
2
2
(xi) Step deviation formula for variance is :
∑d2
∑d2
∑d 2
∑d 2
∑d 2 ∑d
,
(b)
–
,
(c)
– 2 ,
(a) 2 –
n
n
n
n
n
n
(xii) If the distribution is approximately normal, then
3
4
2
(b) M.D. = σ,
(c) M.D. = σ,
(a) M.D. = σ ,
5
5
5
(xiii) For a normal distribution,
2
1
(b) Q.D = σ,
(c) Q.D. = σ,
(a) Q.D. = σ,
3
3
(xiv) For a normal distribution,
(a) Q.D. > M.D.,
(b) Q.D. < M.D.,
(c) Q.D. = M.D.
(xv) For a normal distribution, the range mean ± 1.σ covers
(a) 65%,
(b) 68·26%,
(c) 85%,
Ans. (i) — (c);
(ii) — (a);
(iii) — (b);
(iv) — (b);
(vi) — (a);
(vii) — (b);
(viii) — (a);
(ix) — (a);
(xi) — (b);
(xii) — (c);
(xiii) — (b);
(xiv) — (b);
( )
( )
( )
(d)
(d)
(d)
∑ | X – Md |2
.
n
Q3 – Q1
·
2
( ) ( )
∑d2
n
2
–
∑d
n
2
.
(d) None of these.
(d) None of these.
(d) 95% of the items.
(v) — (a);
(x) — (d);
(xv) — (b).
III. Fill in the blanks :
(i) Algebraic sum of deviations is zero from ……
(ii) The sum of absolute deviations is minimum from ……
(iii) Standard deviation is always …… than range.
(iv) Standard deviation is always …… than mean deviation.
(v) All relative measures of dispersion are …… from units of measurement.
(vi) Variance is the ……value of mean square deviation.
(vii) If Q1 = 10, Q3 = 40, the coefficient of quartile deviation is ……
(viii) If 25% of the items in a distribution are less than 10 and 25% are more than 40, quartile deviation is
……
(ix) The median and s.d. of distribution are 15 and 5 respectively. If each item is increased by 5, the new
median = …… and s.d. = ……
6·52
BUSINESS STATISTICS
(x) A computer showed that the s.d. of 40 observations ranging from 120 to 150 is 35. The answer is
correct/wrong. Tick right one.
Ans. (i) Arithmetic mean (ii) Median (iii) Less (iv) Greater (v) Free
(vi) Minimum
(vii) 0·6
(viii) 15 (ix) 20, 5
(x) Wrong, since s.d. can’t exceed range.
IV. Fill in the blanks :
(i) The …… the Lorenz Curve is from the line of equal distribution, the greater is the variability in the
series.
(ii) Quartile deviation is …… measure of dispersion.
(iii) If Q1 = 20 and Q3 = 50, the coefficient of quartile deviation is ……
(iv) If in a series, coefficient of variation is 64 and mean 10, the standard deviation shall be ……
Ans. (i) Farther
(ii) Absolute
(iii) 0·375 (iv) 6·4.
V. State whether the following statements are true or false. In case of false statements, give the correct
statement.
(i) Algebraic sum of deviations from mean is minimum.
(ii) Mean deviation is least when calculated from median.
(iii) Variance is always non-negative.
(iv) Mean, standard deviation and coefficient of variation have same units.
(v) Relative measures of dispersion are independent of units of measurement.
(vi) Mean and standard deviation are independent of change of origin.
(vii) Variance is square of standard deviation.
(viii) Standard deviation is independent of change of origin and scale.
(ix) Variance is the minimum value of mean square deviation.
2
(x) Q.D. = × (s.d.), always.
3
(xi) Mean deviation can never be negative.
2
(xii) M.D. = × σ, for normal distribution.
3
(xiii) If mean and s.d. of a distribution are 20 and 4 respectively, C.V. = 15%.
(xiv) If each value in a distribution of 5 observations is 10, then its mean is 10 and variance is 1.
(xv) In a discrete distribution s.d. ≥ M.D. (about mean).
Ans.
(i) False,
(ii) True,
(iii) True,
(iv) False,
(v) True,
(vi) False,
(vii) True,
(viii) False,
(ix) True,
(x) False,
(xi) True,
(xii) False,
(xiii) False,
(xiv) False,
(xv) True.
VI. The mean and s.d. of 100 observations are 50 and 10 respectively. Find the new mean and standard deviation,
(i) if 2 is added to each observation.
(ii) if 3 is subtracted from each observation.
(iii) if each observation is multiplied by 5. (iv) if 2 is subtracted from each observation and then it is
divided by 5.
Ans. (i) 52,10,
(ii) 47, 10,
(iii) 250, 50
(iv) 9·6, 2
VII. The sum of squares of deviations of 15 observations from their mean 20 is 240. Find (i) s.d. and (ii) C.V.
Ans. σ = 4, C.V. = 20.
VIII. State, giving reasons, whether the following statements are true or false.
(i) Standard deviation can never be negative
(ii) Sum of squares of deviations measured from mean is least.
Ans. (i) True, (ii) True.
IX. Comment briefly on the following statements :
(i) The mean of the combined series lies between the means of the two component series.
(ii) The standard deviation of the combined series lies between the standard deviations of the two
component series.
(iii) Mean can never be equal to standard deviation.
(iv) Mean can never be equal to variance.
(v) A consistent cricket player has greater variability in test scores.
6·53
MEASURES OF DISPERSION
Ans. (i) True
(ii) False
(iii) False
(iv) False
(v) False.
X. Comment briefly on the following statements :
(i) The median is the point about which the sum of squared deviations is minimum.
—
—
(ii) Since ∑(Xi – X ) = 0, ∴ ∑(Xi – X )2 = 0.
(iii) A computer obtained the standard deviation of 25 observations whose values ranged from 65 to 85 as
25.
(iv) A student obtained the mean and variance of a set of 10 observations as 10, – 5 respectively.
(v) The range is the perfect measure of variability as it includes all the measurements.
(vi) For the distribution of 5 observations :
8, 8, 8, 8, 8,
mean = 8 and variance = 8
(vii) If the mean and s.d. of distribution A are smaller than the mean and s.d. of distribution B respectively,
then the distribution A is more uniform (less variable) than the distribution B.
Ans. (i) False,
(ii) False, (iii) False, (iv) False, (v) False, (vi) False,
(vii) False.
XI.
(a) If s.d. of a group is 15, find the most likely value of (i) Mean deviation and (ii) Quartile deviation of
that group.
(b) If mean deviation of a distribution is 20, find the most likely value of (i) s.d. and (ii) Q.D.
(c) If quartile deviation of a distribution is 6, find the most likely value of (i) s.d. and (ii) M.D.
(d) If quartile deviation of a distribution is 20 and its mean is 60, obtain the most likely value of
(i) Coefficient of variation, (ii) Mean deviation and (iii) Coefficient of mean deviation.
In all the above parts. assume that the distribution is normal.
Ans. (a) M.D. = 12, Q.D. = 10,
(b) s.d. = 25, Q.D. = 16·67,
(c) s.d. = 9, M.D. = 7·2,
(d) C.V. = 50, M.D. = 24, Coefficient of M.D. = 0·4.
Skewness, Moments and
Kurtosis
7
7·1. INTRODUCTION
It was pointed out in the last two chapters that we need statistical measures which will reveal clearly
the salient features of a frequency distribution. The measures of central tendency tell us about the
concentration of the observations about the middle of the distribution and the measures of dispersion give
us an idea about the spread or scatter of the observations about some measure of central tendency. We may
come across frequency distributions which differ very widely in their nature and composition and yet may
have the same central tendency and dispersion. For example, the following two frequency distributions
–
have the same mean X = 15 and standard deviation σ = 6, yet they give histograms which differ very widely
in shape and size.
Frequency distribution I
Class
0— 5
5—10
10—15
15—20
20—25
25—30
Frequency
10
30
60
60
30
10
Frequency distribution II
Class
0— 5
5—10
10—15
15—20
20—25
25—30
Frequency
10
40
30
90
20
10
Thus these two measures viz., central tendency and dispersion are inadequate to characterise a
distribution completely and they must be supported and supplemented by two more measures viz., skewness
and kurtosis which we shall discuss in the following sections. Skewness helps us to study the shape i.e.,
symmetry or asymmetry of the distribution while kurtosis refers to the flatness or peakedness of the curve
which can be drawn with the help of the given data. These four measures viz., central tendency, dispersion,
skewness and kurtosis are sufficient to describe a frequency distribution completely.
7·2. SKEWNESS
Literal meaning of skewness is ‘lack of symmetry’. We study skewness to have an idea about the shape
of the curve which we can draw with the help of the given frequency distribution. It helps us to determine
the nature and extent of the concentration of the observations towards the higher or lower values of the
variable. In a symmetrical frequency distribution which is unimodal, if the frequency curve or histogram is
folded about the ordinate at the mean, the two halves so obtained will coincide with each other. In other
words, in a symmetrical distribution equal distances on either side of the central value will have same
frequencies and consequently both the tails, (left and right), of the curve would also be equal in shape and
length (Fig. 7.1). A distribution is said to be skewed if :
(i) The frequency curve of the distribution is not a symmetric bell-shaped curve but it is stretched more
to one side than to the other. In other words, it has a longer tail to one side (left or right) than to the other. A
frequency distribution for which the curve has a longer tail towards the right is said to be positively skewed
(Fig. 7.2) and if the longer tail lies towards the left, it is said to be negatively skewed (Fig. 7.3).
Figures 7·1 to 7·3 are given on page7·2.
7·2
BUSINESS STATISTICS
(ii) The values of mean (M), median (Md) and mode (Mo) fall at different points i.e., they do not
coincide.
(iii) Quartiles Q1 and Q 3 are not equidistant from the median i.e.,
Q3 – Md ≠ Md – Q1
SYMMETRICAL DISTRIBUTION
M = M o = Md
Fig. 7.1.
POSITIVELY SKEWED
DISTRIBUTION
Mo MdM
Fig. 7.2.
NEGATIVELY SKEWED
DISTRIBUTION
M Md Mo
Fig. 7.3.
Remark. Since the extreme values give longer tails, a positively skewed distribution will have greater
variation towards the higher values of the variable and a negatively skewed distribution will have greater
variation towards the lower values of the variable. For example the distribution of mortality (death) rates
w.r.t. the age after ignoring the accidental deaths will give a positively skewed distribution. However, most
of the phenomena relating to business and economic statistics give rise to negatively skewed distribution.
For instance, the distributions of the quantity demanded w.r.t. the price; or the number of depositors w.r.t.
savings in a bank or the number of persons w.r.t. their incomes or wages in a city will give negatively
skewed curves.
7·2·1. Measures of Skewness. Various measures of skewness (Sk) are :
(1) Sk = Mean – Median = M – Md
…(7·1)
or Sk = Mean – Mode = M – Mo
…(7·1 a)
(2) Sk = (Q3 – Md) – (Md – Q 1 ) = Q3 + Q1 – 2 Md
…(7·2)
These are the absolute measures of skewness and are not of much practical utility because of the
following reasons :
(i) Since the absolute measures of skewness involve the units of measurement, they cannot be used for
comparative study of the two distributions measured in different units of measurement.
(ii) Even if the distributions are having the same units of measurement, the absolute measures are not
recommended because we may come across different distributions which have more or less identical
skewness (absolute measures) but which vary widely in the measures of central tendency and dispersion.
Thus for comparing two or more distributions for skewness we compute the relative measures of
skewness, also commonly known as coefficients of skewness which are pure numbers independent of the
units of measurement. Moreover, in a relative measure of skewness, the disturbing factor of variation or
dispersion is eliminated by dividing the absolute measure of skewness by a suitable measure of dispersion.
The following are the coefficients of skewness which are commonly used.
7·2·2. Karl Pearson’s Coefficient of Skewness. This is given by the formula :
Mean – Mode M – Mo
Sk =
= σ
…(7·3)
s.d.
But quite often, mode is ill-defined and is thus quite difficult to locate. In such a situation, we use the
following empirical relationship between the mean, median and mode for a moderately asymmetrical
(skewed) distribution :
Mo = 3Md – 2M
…(7·4)
Substituting in (7·3), we get
M – (3Md – 2M) 3(M – Md)
Sk =
=
…(7·5)
σ
σ
Remarks 1. Theoretically, Karl Pearson’s Coefficient of Skewness lies between the limits ± 3, but
these limits are rarely attained in practice.
7·3
SKEWNESS, MOMENTS AND KURTOSIS
2. From (7·3) and (7·5), skewness is zero if M = M o = Md. In other words, for a symmetrical
distribution mean, mode and median coincide i.e., M = Md = Mo.
3.
Sk > 0, if M > Md > Mo
or
if Mo < Md < M
…(7·6)
Thus, for a positively skewed distribution, the value of the mean is the greatest of the three measures
and the value of mode is the least of the three measures.
If the distribution is negatively skewed, the inequality in (7·9) is reversed i.e., the inequalities ‘greater
than’ (i.e., >) and ‘less than’ (i.e., <) are interchanged. Thus :
Sk < 0, if M < Md < Mo
or
if Mo > Md > M
…(7·7)
In other words, for a negatively skewed distribution, of the three measures of central tendency viz.,
mean, median and mode, the mode has the maximum value and the mean has the least value.
4. While ‘dispersion’ studies the degree of variation in the given distribution. skewness attempts at
studying the direction of variation. Extreme variations towards higher values of the variable give a
positively skewed distribution while in a negatively skewed distribution, the extreme variations are towards
the lower values of the variable.
5. In Pearson’s coefficient of skewness, the disturbing factor of variation is eliminated by dividing the
absolute measure of skewness M – Mo by the measure of dispersion σ (standard deviation).
6. If the distribution is symmetrical, say, about mean, then
⇒
Also
Mean – Xmin = Xmax – Mean
Xmax + Xmin = 2 Mean
Xmax – Xmin = Range
…(7·8)
…(7.8a)
…(7.9)
Adding and subtracting (7.8a) and (7.9), we get respectively :
1
Xmax = Mean + Range
…(7.10)
2
1
2 Xmin = 2 Mean – Range
⇒
Xmin = Mean – Range
…(7.10a)
2
Example 7·1. A distribution of wages paid to workers would show that, although a few reach very high
levels, most workers are at lower level of wages. If you were an employer, resisting worker’s claim for an
increase of wages, which average would suit your case? Do you think your argument will be different if you
are a trade union leader ? Explain.
[Delhi Univ. B.A. (Econ. Hons.), 1997]
Solution. The distribution of wages paid to the workers would show that, although a few reach very
high levels, most workers are at lower level of wages. This implies that the distribution of wages of workers
in the factory is positively skewed so that
Mean > Median > Mode
⇒
Mode < Median < Mean.
Since mode is the least of the three averages mean, median and mode, the average that will suit the
employer most (to resist workers claim for higher wages) will be mode.
By similar argument, since mean is the largest of the three averages, it will suit most to the trade union
leader.
Example 7.2. In a symmetrical distribution the mean, standard deviation and range of marks for a
group of 20 students are 50, 10 and 30 respectively. Find the mean marks and standard deviation of marks,
if the students with the highest and the lowest marks are excluded.
[Delhi Univ. B.Com. (Hons.), 2004]
Solution. We are given a symmetrical distribution in which,
2 X max = 2 Mean + Range
⇒
—
No. of observations (n) = 20 ; Mean ( X ) = 50 ; s.d. (σ) = 10
Since the distribution is symmetrical (about mean) , we have
⇒
Mean – Xmin = Xmax – Mean
Xmax + Xmin = 2 Mean = 2 × 50 = 100
and
Range = 30
[From (*)]
…(*)
…(**)
7·4
BUSINESS STATISTICS
Also
Xmax – Xmin = Range = 30 (Given)
…(***)
Adding and subtracting (**) and (***), we get respectively
130
2 Xmax = 130
⇒
Xmax =
= 65
and
2 Xmin = 70
2
Thus the given problem reduces to :
⇒
Xmin =
70
= 35
2
—
“Given n = 20, X = 50 and σX = 10, obtain the mean and standard deviation if two observations 65 and
35 are omitted.”
— ∑X
—
X=
⇒
∑X = nX = 20 × 50 = 1,000
n
—
σ2 = 10 2 = 100
⇒
∑X2 = n (σ2 + X 2 ) = 20 (100 + 2,500) = 52,000
On omitting the two observations 35 and 65, for the remaining, N = n – 2 = 20 – 2 = 18
observations, we have
18
∑ Xi = ∑X – (35 + 65) = 1,000 – 100 = 900
i=1
18
∑ Xi2 = ∑X2 – (352 + 65 2 ) = 52,000 – (1,225 + 4,225) = 46,550
i=1
—
∴
New Mean ( X ′) =
and
New s.d. (σ ′) =
—
1 18
900
∑ X =
= 50 = X
18 i = 1 i 18
1 18 2 — 2
∑ X – ( X ′) =
18 i = 1 i
46‚550
– 50 2 =
18
1‚550
=⎯
√⎯⎯⎯
86·11 = 9·28
18
Aliter : For a symmetrical distribution (about mean) we have, [From (7·10) and (7·10a)],
1
1
Xmax = Mean + Range and Xmin = Mean – Range
2
2
Example 7·3. Calculate Karl Pearson’s co-efficient of skewness from the following data :
Size
Frequency
:
:
1
10
2
18
3
30
4
25
5
12
6
3
7
2
Solution.
Mean (M) =
S.D. (σ) =
=
COMPUTATION OF MEAN, MODE AND S.D.
Frequency (f)
f.x
f.x2
Size (x)
∑ fx 328
= 100 = 3·28
N
( ∑Nfx )
– (
)
∑
–
N
fx2
1258
100
328
100
———————————————————————————————————————————————————————
—————————————————————
2
2
= ⎯√⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
12·58 – 10·7584 = √
⎯⎯⎯⎯⎯⎯
1·8216
= 1·3497
1
2
3
4
5
6
7
Total
10
18
30
25
12
3
2
N = 100
10
36
90
100
60
18
14
∑ fx = 328
10
72
270
400
300
108
98
∑ fx2 = 1258
Since the maximum frequency is 30, corresponding value of x viz., 3 is the mode. Thus Mode (Mo) = 3.
Karl Pearson’s Coefficient of Skewness is given by
M – Mo 3·28 – 3·00 0·28
Sk =
= 1·3497 = 1·3497 = 0·2075
σ
Hence, the distribution is slightly positively skewed.
7·5
SKEWNESS, MOMENTS AND KURTOSIS
Example 7·4. Calculate Karl Pearson’s Coefficient of Skewness from the data given below :
Hourly wages (Rs.)
No. of workers
Hourly wages (Rs.)
No. of workers
40—50
50—60
60—70
70—80
80—90
5
6
8
10
25
90—100
100—110
110—120
120—130
130—140
30
36
50
60
70
Solution
COMPUTATION OF MEAN, MEDIAN AND S.D.
Hourly wages
(Rs.)
40—50
50—60
60—70
70—80
80—90
90—100
100—110
110—120
120—130
130—140
Mid-value (X)
No. of workers
(f)
– 85
d = X 10
–4
–3
–2
–1
0
1
2
3
4
5
5
6
8
10
25
30
36
50
60
70
45
55
65
75
85
95
105
115
125
135
N = ∑f = 300
fd
fd2
–20
–18
–16
–10
0
30
72
150
240
350
80
54
32
10
0
30
144
450
960
1750
∑fd = 778
‘Less than’
c.f.
5
11
19
29
54
84
120
170
230
300
∑fd2 = 3510
Since the maximum frequency viz., 70 occurs towards the end of the frequency distribution, mode is
ill-defined in this case. Hence, we obtain Karl Pearson’s coefficient of skewness using median viz., by the
formula :
Sk =
3 (Mean – Median)
σ
Mean = A +
…(*)
h∑fd
× 778
= 85 + 10300
= 85 + 25·93 = Rs. 110·93
N
∑fd2
N –
s.d. (σ) = h .
( )
∑fd
N
2
= 10 ×
3510
300 –
( )
778
300
2
= 10 × ⎯√⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
11·7 – (2·5932)2 = 10 × ⎯
√⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
11·7 – 6·7252
= 10 × ⎯√⎯⎯⎯⎯
4·9748 = 10 × 2·23043 = Rs. 22·3043
N
2
= 300
2 = 150. The c.f. just greater than 150 is 170. Hence, the corresponding class 110—120 is
the median class. Using the median formula, we get
Median = l +
h
f
( – c ) = 110 +
N
2
10
50 (150
– 120) = 110 + 30
5 = 116
Substituting these values in (*), we get
Sk =
3 (110·93 – 116)
3 × 5·07
–15·21
= – 22·3043 = 22·3043 =
22·3043
Hence, the given distribution is negatively skewed.
– 0·6819
7·6
BUSINESS STATISTICS
Example 7·5. Consider the following distributions :
Distribution A
Mean
100
Median
90
Standard Deviation
10
Distribution B
90
80
10
(i) Distribution A has the same degree of the variation as distribution B.
(ii) Both distributions have the same degree of skewness. True/False ? Comment, giving reasons.
[Delhi Univ. B.Com. (Hons.), 2008]
Solution.
(i)
C.V. for distribution A = 100 ×
σA
–
XA
C.V. for distribution B = 100 ×
σB
–
XB
10
= 100 × 100
= 10
= 100 ×
10
90
= 11·11
Since C.V. (B) > C.V. (A), the distribution B is more variable than the distribution A. Hence, the given
statement that the distribution A has the same degree of variation as distribution B is wrong.
(ii) Karl Pearson’s coefficient of skewness for the distributions A and B is given by :
3(M – Md) 3(100 – 90)
Sk (A) =
=
= 3 ; Sk (B) = 3(9010– 80) = 3
10
σ
Since Sk(A) = Sk(B) = 3, the statement that both the distributions have the same degree of skewness is
true.
Example 7·6. (a) In a moderately asymmetrical distribution, the mode and mean are 32·1 and 35·4
respectively. Calculate the median.
(b) From a moderately skewed distribution of retail prices for men’s shirts, it is found that the mean
price is Rs. 200 and the median price is Rs. 170. If the coefficient of variation is 20%, find the Pearsonian
coefficient of skewness of the distribution. [Delhi Univ. B.Com. (Hons.), 2009; B.A. (Econ. Hons.), 2008]
Solution (a) For a moderately asymmetrical distribution, we have
1
Mo = 3Md – 2M
⇒
3Md = Mo + 2M
⇒
Md = (Mo + 2M )
3
1
1
102·9
∴
Md = (32·1 + 2 × 35·4) = (32·1 + 70·8) =
= 34·3
3
3
3
(b) We are given :
Mean (M) = Rs. 200,
Median (Md) = Rs. 170
100 σ
20M
20 × 200
and
C.V. = 20%
⇒
= 20
⇒
σ=
= Rs.
= Rs. 40.
M
100
100
3 (M– Md) 3 (200 – 170) 9
Sk (Karl Pearson) =
=
= = 2·25
σ
40
4
Example 7·7. Pearson’s coefficient of skewness for a distribution is 0·4 and its coefficient of variation
is 30%. Its mode is 88, find mean and median.
[Delhi Univ. B.Com. (Hons.), 1997]
Solution. We are given Mode = Mo = 88, Karl Pearson’s coefficient of skewness is :
M – Mo M – 88
Sk =
=
= 0·4 (given)
…(*)
σ
σ
σ
C.V. = 100
M = 30 (given)
σ = 30M
100 = 0·3M
⇒
…(**)
From (*) and (**), we get
M – 88 = 0·4σ = 0·4 × 0·3 M
Substituting in (**), we get
⇒
M – 0·12M = 88
σ = 0·3 × 100 = 30
⇒
88
(1 – 0·12) M = 88 ⇒ M = 0·88
= 100
7·7
SKEWNESS, MOMENTS AND KURTOSIS
Using the empirical relation between mean, median and mode for a moderately asymmetrical
distribution viz.,
Mo = 3Md – 2M
⇒
3Md = Mo + 2M,
we get :
Md = 31 (Mo + 2M) = 31 (88 + 2 × 100) = 288
3 = 96
Hence, Mean = 100 and Median = 96.
Example 7·8. Pearson’s measure of skewness of a distribution is 0·5. Its median and mode are
respectively 42 and 36. Find the Coefficient of Variation.
Solution. We are given : Median = 42, Mode = 36
⇒
and Pearson’s coefficient of skewness = 0·5
…(*)
Sk =
Mean – Mode
=
σ
0·5
…(**)
To find s.d. (σ), we shall first find the value of mean, by using the empirical relationship between
mean, median and mode for a moderately asymmetrical distribution viz.,
3 Median – Mode 3 × 42 – 36 126 – 36 90
Mode= 3 Median – 2 Mean
⇒
Mean =
=
= 2 = 2 = 45
2
2
Mean – Mode 45 – 36 9
Substituting in (**), we get :
(s.d.) σ =
= 0·5 = 0·5 = 18
0·5
× σ 100 × 18
Coefficient of Variation (C.V.) = 100
= 40
Mean =
45
Hence, C.V. is 40.
Example 7·9. The sum of 50 observations is 500 and the sum of their squares is 6,000 and median is
12. Compute the coefficient of variation and the coefficient of skewness. [Delhi Univ. B.Com. (Pass), 2000]
Solution. In the usual notations, we are given :
∴
n = 50 , ∑x = 500
and
∑x2 = 6,000 ; Median = 12
∑x 500
∑x2 – 2 6‚000
Mean ( –x ) =
= 50 = 10 ; σ2 =
– x = 50 – 100 = 20 ⇒
n
n
σ = √⎯⎯20 = 4·47
σ 100 × 4·47
Coefficient of Variation (C.V.) = 100
= 44·7
– =
10
x
Karl Pearson’s coefficient of skewness is given by :
3(Mean – Median) 3(10 – 12) – 6
Sk =
= 4·47 = 4·47 = –1·34.
σ
Example 7·10. The following information was obtained from the records of a factory relating to the
wages.
Arithmetic Mean: Rs. 56·80; Median : Rs. 59·50;
S.D. : Rs. 12·40
Give as much information as you can about the distribution of wages.
Solution. We can obtain the following information from the above data :
(i) Since Md = Rs. 59·50, we conclude that 50% of the workers in the factory obtain the wages above
Rs. 59·50.
σ 100 × 12·40
(ii)
C.V. = 100
= 21·83
M =
56·80
(iii) Karl Pearson’s coefficient of skewness is given by :
3(M – Md) 3(56·80 – 59·50) 3 × (–2·70) – 8·1
Sk =
=
= 12·40 = 12·4 = – 0·65
12·40
σ
Hence, the distribution of wages is negatively skewed i.e., it has a longer tail towards the left.
(iv) Using the empirical relation between M, Md and Mo for a moderately asymmetrical distribution,
we get
Mo = 3Md – 2M = 3 × 59·50 – 2 × 56·80 = 178·50 – 113·60 = Rs. 64·90.
7·8
BUSINESS STATISTICS
Example 7·11. You are given the position in a factory before and after the settlement of an industrial
dispute. Comment on the gains or losses from the point of view of workers and that of management.
Before
After
No. of workers
3,000
2,900
Mean wage (in Rs.)
220
230
Median wage (in Rs.)
250
240
Standard deviation (in Rs.)
30
26
[Delhi Univ. B.Com. (Hons.), 2006]
Solution. On the basis of the above data we are in a position to make the following comments :
(i) The number of workers after the dispute has decreased from 3000 to 2900. Obviously this is a
definite loss to the persons thrown out or retrenched. It may also be a loss to the management if their
retrenchment affects the efficiency of work adversely.
Total wages paid
(ii) We know that :
Average wage =
Total No. of workers
⇒
Total wages paid = (Average wage) × (Total No. of workers)
Hence,
Total wages paid by the management before the dispute = Rs. 3000 × 220 = Rs. 6,60,000
Total wages paid by the management after the dispute
= Rs. 2900 × 230 = Rs. 6,67,000
Thus we see that the total wages paid by the management have gone up after the dispute (the additional
wage bill being Rs. 7000), although the number of workers has been reduced from 3000 to 2900. This is
due to the fact that the average wage per worker has increased after the dispute - which is a distinct
advantage to the workers.
It may be pointed out that the increased wages paid by the management (Rs. 7000) should not be
viewed as a disadvantage to the management unless we have definite reasons to believe that the efficiency
and productivity have not gone up after the dispute. However, the loss to the managements due to higher
wage bill, will be more than compensated if after the dispute, there is a definite increase in the efficiency of
the workers or/and increase in productivity.
(iii) Although the number of workers has decreased from 3000 to 2900 after the dispute, the average
wage per worker has gone up from Rs. 220 to 230. This might probably be a consequence of the
retrenchment of casual labour or temporary labour working on daily wages or so with relatively lower
wages.
(iv) The median wage after the dispute has come down from Rs. 250 to Rs. 240. This implies that
before the dispute upper 50% of the workers were getting wages above Rs. 250 whereas after the dispute
they get wages only above Rs. 240.
(v) Using the empirical relation between mean, median and mode (for a moderately asymmetrical
distribution) viz.,
Mode = 3 Median – 2 Mean
We get
Mode (before dispute) = 3 × 250 – 2 × 220 = Rs. 310
Mode (after dispute) = 3 × 240 – 2 × 230 = Rs. 260
Thus, we find that the modal wage has come down from Rs. 310 (before dispute) to Rs. 260 (after
dispute). Thus after the dispute there is concentration of wages around a much smaller value.
× (s.d.)
C.V. = 100Mean
(vi)
∴ C.V. (before dispute) =
100 × 30
220 =
13·64
and
× 26
C.V. (after dispute) = 100230
= 11·30
Since C.V. has decreased from 13·64 to 11·30, the distribution of wages has become less variable i.e.,
more consistent or uniform after the settlement of the dispute. Thus, after the settlement there are less
disparities in wages and from management point it will result in greater satisfaction to the workers.
7·9
SKEWNESS, MOMENTS AND KURTOSIS
(vii) Since we are given mean and median, we can calculate Karl Pearson’s coefficient of skewness for
studying the symmetry of the distribution of wages before the dispute and after the dispute viz.,
Sk = 3(M – Md))/σ
Sk (before dispute) = 3(22030– 250) = –3
and Sk (after dispute) = 3(23026– 240) = –1·15
Thus the highly negatively skewed distribution (before the dispute) has become a moderately
negatively skewed distribution (after the dispute). This implies that the curve of distribution of wages after
the dispute has a less longer tail towards the left. In other words, the number of workers getting lower
wages has increased.
EXERCISE 7·1
1. (a) Explain the concept of skewness. Draw the sketch of a skewed frequency distribution and show the position
of the mean, median and mode when the distribution is asymmetric.
(Delhi Univ. B.Com., 1997)
(b) Explain the concept of positive and negative skewness.
(c) Show graphically the positions of mean, median and mode in a positively and negatively skewed series.
[Delhi Univ. B.Com. (Pass), 1998]
2. Comment on the following :
(a) In a symmetrical distribution, we have mean = median ≠ mode.
(b) “A series representing U-shaped curve is symmetrical.” Comment.
[Delhi Univ. B.Com. (Hons.) 2002]
[Delhi Univ. B.Com. (Pass), 2001]
3. (a) The values of mean and median are 30 and 40 respectively in a frequency distribution. Is the distribution
skewed? If yes, state the direction of skewness.
[Delhi Univ. B.Com. (Pass), 2002]
Ans. Mean < Median, the distribution is negatively skewed.
(b) The mean for a symmetrical distribution is 50·6. Find the values of median and mode.
[C.S. (Foundation), June 2001]
Ans. Median = Mode = 50·6.
4. (a) Define Pearson’s measure of skewness. What is the difference between relative measure and the absolute
measure of skewness.
(b) From the following data find out Karl Pearson’s co-efficient of skewness :
Measurement
:
10
11
12
13
14
15
Frequency
:
2
4
10
8
5
1
Ans. 0·3478.
5. Calculate the Pearson’s coefficient of skewness from the following :
Wages (Rs.)
:
0—10
10—20
20—30
No. of Workers
:
15
20
30
30—40
25
40—50
10
Ans. 0·1845.
6. Calculate Karl Pearson’s coefficient of skewness from the following data and explain its significance :
Wages
:
70—80
80—90
90—100 100—110 110—120 120—130 130—140 140—150
No. of Persons :
12
18
35
42
50
45
20
8
[Delhi Univ. B.Com. (Hons.), 2000]
Ans. M = Rs. 110·43, Mo = Rs. 116·15, σ = Rs. 17·26, Sk = – 0·3316.
7. The mean, standard deviation and range of a symmetrical distribution of weights of a group of 20 boys are 40
kgs., 5 kgs. and 6 kgs. respectively. Find the mean and standard deviation of the group if the lightest and heaviest boys
are excluded.
[Delhi Univ. B.A. (Econ. Hons.), 2004]
Ans. Mean = 40 , s.d. (σ) = 5.17
Hint. Proceed as in Example 7.2.
8. Calculate the Pearson’s measure of skewness on the basis of Mean, Mode and Standard Deviation.
Mid-value (X)
:
14.5
15.5
16.5
17.5
18.5
19.5
20.5
f
:
35
40
48
100
125
87
43
Hint. The corresponding classes are : 14—15, 15—16, …, 21—22.
Ans. Mean = 18.07 , Mode = 18.4, σ = 1.775, Skewness (Pearson) = – 0.186
21.5
22
7·10
BUSINESS STATISTICS
9. From the following data of age of employees, calculate coefficient of skewness and comment on the result :
Age below (yrs.)
:
25
30
35
40
45
50
55
No. of employees :
8
20
40
65
80
92
100
[Delhi Univ. MBA, 1997]
–
Ans. x = 37·25 yrs.; Mo = 36·67 yrs.; σ = 16·99 yrs.; Sk (Karl Pearson) = 0·07.
10. Calculate Karl Pearson’s coefficient of skewness from the following series :
Wt. in kgs.
:
Below 40
40—50
50—60
No. of persons
:
10
16
18
Wt. in kgs.
:
80—90
90—100
100 and above
No. of persons
:
4
4
3
60—70
25
70—80
20
Hint. Take the first class as 30—40 and the last class as 100—110.
Ans. Mean = 62·200 kg,
Mode = 65·833 kg,
σ = 16·857 kg., Sk = – 0·2155.
11. Calculate Karl Pearson’s coefficient of skewness from the following data :
Marks (above)
:
0
10
20
30
40
50
60
70
80
No. of students
:
150
140
100
80
80
70
30
14
0
Hint. Locate mode by the method of grouping ; two modal classes 10—20 and 50—60; mode ill-defined. Find
median.
Ans. Sk = 3(M – Md)/σ = – 0·6622.
12. The daily expenditure of 100 families is given below :
Daily Expenditure :
0—20
20—40
40—60
60—80
No. of Families
:
13
?
27
?
If the mode of the distribution is 44, calculate the Karl Pearson coefficient of skewness.
Ans. Frequency for the class 20—40 is 25 and for the class 60—80 is 19.
Sk (Karl Pearson) =
50 – 44
25·3 =
80—100
16
0·24.
13. The following facts are gathered before and after an industrial dispute :
Before dispute
No. of workers employed
515
Mean wages
Rs. 49·50
Median wages
Rs. 52·80
Variance of wages
(Rs.)2 121·00
Compare the position before and after the dispute in respect of
(a) total wages,
(b) modal wages,
(c) standard deviation,
Ans.
Before dispute
(i) Total wages
Rs. 25,492·50
(ii) Modal wages
Rs. 59·40
(iii) C.V.
22·22
(iv) Skewness
– 0·90
After dispute
509
Rs. 52·70
Rs. 50·00
(Rs.)2 144·00
and
(d) skewness.
After dispute
Rs. 26,849·75
Rs. 44·50
22·74
0·69
14. You are given the position in a factory before and after the settlement of an industrial dispute.
Before Dispute
After Dispute
No. of workers
3,000
2,900
Mean wages (Rs.)
220
230
Median wages (Rs.)
250
240
Standard deviation (Rs.)
30
26
Comment on the gains and losses from the point of view of workers and that of management.
[Delhi Univ. B.Com. (Hons.), 2006]
Hint. Proceed as in Example 7.13.
15. You are given below the following details relating to the wages is respect of two factories from which it is
concluded that the skewness and variability are same in both the factories.
SKEWNESS, MOMENTS AND KURTOSIS
7·11
Factory A
Factory B
Arithmetic Mean :
50
45
Mode
:
45
50
Variance
:
100
100
Point out the mistake or the wrong inference in the above statement.
Ans. C.V. (A) = 20, C.V. (B) = 22·2 ; Sk(A) = +0·5, Sk(B) = – 0·5.
16. The sum of 20 observations is 300 and its sum of squares is 5,000 and median is 15. Find the coefficient of
skewness and coefficient of variation.
[Delhi Univ. B.Com. (Hons.), 2007]
5
Ans. Sk = 0, C.V. = 15
× 100 = 33·3.
17. For a group of 10 items ∑X = 452, ∑X 2 = 24,270 and Mode = 43·7. Find the Pearsonian coefficient of
skewness.
Ans. Sk = 0·08.
18. If the mode and mean of a moderately asymmetrical series are respectively 16 inches and 15·6 inches, would
be its most probable median ?
Ans. Median = 15·73 inches.
19. In a slightly skew distribution the arithmetic mean is Rs. 45 and the median is Rs. 48. Find the approximate
value of mode.
Ans. Mode = Rs. 54.
20. In a frequency distribution, Karl Pearson’s coefficient of skewness revealed that the distribution was skewed to
the left to an extent of 0·6. Its mean value was less than its modal value by 4·8. What was the standard deviation ?
Ans. σ = 8.
21. In a distribution mean = 65, median = 70 and the coefficient of skewness is – 0·6. Find mode and coefficient of
variation. (Assume that the distribution is moderately asymmetrical.)
Ans. Mode = 80, C.V. = 38·46
22. In a certain distribution the following results were obtained :
–
Arithmetic Mean (X) = 45; Median = 48; Coefficient of skewness = – 0·4
The person who gave you this data, failed to give you S.D. (Standard Deviation). You are required to estimate it
with the help of the above data.
[Delhi Univ. B.Com. (Hons.), 1997]
Ans. σ = 22·5.
23. Karl Pearson’s measure of skewness of a distribution is 0·5. The median and mode of the distribution are
respectively, 42 and 32.
Find : (i) Mean, (ii) the S.D., (iii) the coefficient of variation.
[Delhi Univ. B.A. (Econ. Hons.), 2000]
Ans. (i) 47, (ii) 30, (iii) 63·83.
24. Karl Pearson’s coefficient of skewness of a distribution is 0·32. Its standard deviation is 6·5 and mean is 29·6.
Find the mode and median of the distribution.
If the mode of the above is 24·8, what will be the standard deviation ?
[Delhi Univ. B.Com. (Hons.), 1998]
Ans. Mode = 27·52, Median = 28·91, σ = 15.
25. Karl Pearson’s coefficient of skewness of a distribution is +0·40. Its standard deviation is 8 and mean is 30.
Find the mode and median of the distribution.
Ans. Mode = 26·8, Median = 28·93.
26. The median, mode and coefficient of skewness for a certain distribution are respectively 17·4, 15·3 and 0·35.
Calculate the coefficient of variation.
Ans. Mean = 18·45; C.V. = 48·78.
27. Karl Pearson’s coefficient of skewness of a distribution is 0.5. Its mean is 30 and coefficient of variation is
20%. Find the mode and median of the distribution.
[Delhi Univ. B.A. (Econ. Hons.), 2007]
Ans. σ = 6, Median = 29, Mode = 27.
28. (a) A frequency distribution is positively skewed. The mean of the distribution is :
Greater than the mode, Less than the mode, Equal to the mode, None of these.
Tick the correct answer.
7·12
BUSINESS STATISTICS
(b) In a moderatly skewed distribution the values of mean and median are 5 and 6 respectively. The value of mode
in such a situation is approximately equal to… .
(i) 8
(ii) 11
(iii) 16
(iv) None of these.
Ans. (a) M > Mo, (b) : (i) 8.
29. State the empirical relationship among mean, median and mode in a symmetrical and moderately asymmetrical
frequency distribution. How does it help in estimating mode and measuring skewness ?
Given that mean of a distribution in 50 and mode in 58;
(i) Calculate the median;
(ii) What can you say about the shape of the distribution ? Explain, Why ? [Delhi Univ. B.A.. (Econ. Hons.), 2009]
Ans. (i) Md = 52.67; (ii) M < Md < M0
⇒ Distribution is negatively skewed.
30. In a moderately asymmetrical distribution :
(i) The mode and median are 300 and 240 respectively. Find the value of mean.
(ii) Mean = 200 and Mode = 150. Find the value of median.
[C.S. (Foundation), Dec. 2001]
Ans. (i) Mean = 210, (ii) Median = 183·33.
31. Which group is more skewed ?
(i) Mean = 22 ; Median = 24 ; s.d. = 10. (ii) Mean = 22 ; Median = 25 ; s.d. = 12.
Ans. Sk (i) = – 0·60 ; Sk (ii) = – 0·75. Group (ii) is more skewed to the left.
32. What is the relationship between mean, mode and median ? What is the condition under which this relationship
holds ? Locate graphically the position of the three measures in the case of both negatively as well as positively skewed
distribution.
Ans. Mode = 3 Median – 2 Mean, for a moderately asymmetrical distribution.
7·2·3. Bowley’s Coefficient of Skewness. Prof. A.L. Bowley’s coefficient of skewness is based on the
quartiles and is given by :
(Q – Md) – (Md – Q 1 )
Q + Q1 – 2Md
Sk = 3
⇒
Sk = 3
…(7·11)
(Q 3 – Md) + (Md – Q1)
Q3 – Q1
Remarks 1. Bowley’s coefficient of skewness is also known as Quartile coefficient of skewness and is
especially useful in situations where quartiles and median are used viz., :
(i) When the mode is ill-defined and extreme observations are present in the data.
(ii) When the distribution has open end classes or unequal class intervals.
In these situations, Pearson’s coefficient of skewness cannot be used.
2. From (7·11), we observe that :
Sk = 0, if Q3 – Md = Md – Q 1
…(7·12)
This implies that for a symmetrical distribution (Sk = 0), median is equidistant from the upper and
lower quartiles. Moreover, skewness is positive if :
Q3 – Md > Md – Q 1
⇒
Q3 + Q1 > 2Md
…(7·13)
and skewness is negative if :
Q3 – Md < Md – Q 1
⇒
Q3 + Q1 < 2Md
…(7·14)
3. Limits for Bowley’s Coefficient of Skewness.
Sk (Bowley) ⎟ ≤1
⇒
–1 ≤ Sk (Bowley) ≤ 1.
…(7·15)
Thus, Bowley’s coefficient of skewness ranges from –1 to 1.
Further, we note from (7·11) that :
Sk = +1, if Md – Q1 = 0,
i.e.,
if Md = Q1
…(7·15a)
and
Sk = –1, if Q3 – Md = 0,
i.e.,
if Q3 = Md
…(7·15b)
4. It should be clearly understood that the values of the coefficient of skewness obtained by Bowley’s
formula and Pearson’s formula are not comparable, although in each case, Sk = 0 implies the absence of
skewness i.e., the distribution is symmetrical. It may even happen that one of them gives positive skewness
while the other gives negative skewness. (See Example 7·17)
7·13
SKEWNESS, MOMENTS AND KURTOSIS
5. The only and perhaps quite serious limitations of this coefficient is that it is based only on the central
50% of the data and ignores the remaining 50% of the data towards the extremes.
7·2·4. Kelly’s Measure of Skewness. The drawback of Bowley’s coefficient of skewness, (viz., that it
ignores the 50% of the data towards the extremes), can be partially removed by taking two deciles or
percentiles equidistant from the median value. The refinement was suggested by Kelly. Kelly’s percentile
(or decile) measure of skewness is given by :
Sk = (P 90 – P50) – (P50 – P10) = P90 + P10 – 2P50
…(7·16)
But P 90 = D9 and P10 = D1. Hence, (7·16) can be re-written as :
Sk = (D 9 – D5) – (D5 – D1) = D9 + D1 – 2D5
…(7·16a)
Pi and D i are the ith percentile and decile respectively of the given distribution.
(7·16) or (7·16a) gives an absolute measure of skewness. However, for practical purposes, we
generally compute the coefficient of skewness, which is given by :
Sk (Kelly) =
=
(P90 – P50) – (P50 – P10)
(P90 – P50) + (P50 – P10)
=
P90 + P10 – 2P50
P90 – P10
…(7·17)
D9 + D1 – 2D5
D9 – D1
…(7·17a)
Remarks 1. We have : D5 = P50 = Median. Hence, from (7·17) and (7·17a), we get
Sk (Kelly) =
P90 + P10 – 2Md
P90 – P10
=
D9 + D1 – 2Md
D9 – D1
…(7·18)
2. This method is primarily of theoretical importance only and is seldom used in practice.
7·2·5. Coefficient of Skewness based on Moments. This coefficient is based on the 2nd and 3rd
moments about mean and is discussed in detail in § 7·5.
Example 7·12. Comment on the following :
(i) The mode of a distribution cannot be less than arithmetic mean.
(ii) If Q1 , Q2, Q3 be respectively the lower quartile, the median and the upper quartile of a
distribution, then Q2 – Q1 = Q3 – Q2.
Solution. (i) The given statement is not true in general. Mode of a distribution cannot be lower than
arithmetic mean or in other words, arithmetic mean is greater than mode only for a positively skewed
distribution. However, Mean = Mode for a symmetrical distribution, while for a negatively skewed
distribution Mean < Mode.
(ii) The statement : Q3 – Q2 = Q2 – Q1,
is true only for a symmetrical distribution. However, if the distribution is skewed, then
Q3 – Q2 ≠ Q2 – Q1
If Q3 – Q2 > Q2 – Q1 ⇒
Q1 + Q3 > 2Q2, the distribution is positively skewed.
and if Q3 – Q2 < Q2 – Q1 ⇒
Q1 + Q3 < 2Q2, the distribution is negatively skewed.
Example 7·13. Calculate Bowley’s coefficient of skewness of the following data :
Weight (lbs)
Under 100
100—109
110—119
120—129
No. of persons
1
14
66
122
Weight (lbs)
No. of persons
Weight (lbs)
No. of persons
130—139
140—149
150—159
160—169
145
121
65
31
170—179
180—189
190—199
200 and over
12
5
2
2
7·14
BUSINESS STATISTICS
COMPUTATION OF QUARTILES
Solution. Here we are given the frequency Weight
No. of persons ‘Less than’ Class boundaries
(f)
c.f.
(lbs)
distribution with inclusive type classes. Since the
formulae for median and quartiles are based on
0—99
1
1
0 — 99·5
continuous frequency distribution with exclusive
100—109
14
15
99·5—109·5
type classes without any gaps, we obtain the
66
81
109·5—119·5
class boundaries which are given in the last 110—119
120—129
122
203
119·5—129·5
column of the adjoining table.
————————————————
——————————————————————
————————————————
——————————————————————————
3N
Here N = 586, 4N = 146·5, N
2 = 293, 4 = 439·5
The c.f. just greater than N/2 i.e., 293 is 348.
Hence, the corresponding class 129·5—139·5 is
the median class. Using the Median formula, we
get
10
Md (Q2) = 129·5 + 145
(293 – 203)
130—139
140—149
150—159
160—169
170—179
180—189
190—199
200—209
145
121
65
31
12
5
2
2
348
469
534
565
577
582
584
N = 586
129·5—139·5
139·5—149·5
149·5—159·5
159·5—169·5
169·5—179·5
179·5—189·5
189·5—199·5
199·5—209·5
× 90
= 129·5 + 10145
= 129·50 + 6·21 = 135·71
The c.f. just greater than N/4 i.e., 146·5 is 203. Hence, the corresponding class 119·5—129·5 contains Q1
∴
10
× 65·5
Q1 = 119·5 + 122
(146·5 – 81) = 119·5 + 10 122
= 119·50 + 5·37 = 124·87
The c.f. just greater than 3N/4 i.e., 439·5 is 469. Hence, the corresponding class 139·5—149·5 contains Q3.
∴
10
× 91·5
Q3 = 139·5 + 121
(439·5 – 348) = 139·5 + 10 121
= 139·50 + 7·56 = 147·06
Bowley’s coefficient of skewness is given by :
Sk (Bowley) =
Q3 + Q1 – 2Md
Q3 – Q1
+ 124·87 – 2 × 135·71 271·93 – 271·42
0·51
= 147·06147·06
=
= 22·19
= 0·0298.
– 124·87
22·19
Example 7·14. Calculate the coefficient of skewness from the following data by using quartiles.
Marks
No. of students
Marks
No. of students
Above 0
180
Above 60
65
Above 15
160
Above 75
20
Above 30
130
Above 90
5
Above 45
100
COMPUTATION OF QUARTILES
Number of students (f)
Marks
Solution. We are given ‘more than’ cumulative
frequency distribution. To compute quartiles, we first
express it as a continuous frequency distribution
without any gaps as given in the adjoining table.
N 180
2= 2 =
90,
N 180
4= 4 =
45 and
3N
4 =
∴
Md = l +
h
f
(
N
2 –C
0—15
15—30
30—45
45—60
60—75
75—90
above 90
135
The c.f. just greater than N/2 i.e., 90 is 115. Hence the
corresponding class 45—60 is the median class.
‘Less than’ c.f.
——————————————————————————————————————————————————————————————————————
180 – 160 = 20
160 – 130 = 30
130 – 100 = 30
100 – 65 = 35
65 – 20 = 45
20 – 5 = 15
5
20
50
80
115
160
175
180
——————————————————
———————————————————————————————————————————————————————
) = 45 + ( 90 – 80 ) = 45 +
15
35
Total
15 × 10
35 =
N = ∑f = 180
45 + 4·29 = 49·29
The c.f. just greater than N/4 i.e., 45 is 50. Hence, the corresponding class 15—30 contains Q1.
∴
Q1 = l + hf
(
N
4 –C
) = 15 + ( 45 – 20 ) = 15 +
15
30
15 × 25
30 =
15 + 12·5 = 27·5
The c.f. just greater than 3N/4 i.e., 135 is 160. Hence, the corresponding class 60—75 contains Q3.
7·15
SKEWNESS, MOMENTS AND KURTOSIS
Q3 = l + hf
∴
(
3N
4 –C
) = 60 + ( 135 – 115 ) = 60 +
15
45
20
=
3
60 + 6·67 = 66·67
Hence, Bowley’s coefficient of skewness based on quartiles is given by :
Q + Q1 – 2Md 66·67 + 27·50 – 2 × 49·29 94·17 – 98·58
4·41
Sk (Bowley) = 3
=
=
= – 39·17
= – 0·1126
66·67 – 27·50
39·17
Q3 – Q1
Example 7·15. If the first quartile is 142 and the semi-interquartile range is 18, find the median
(assuming the distribution to be symmetrical.
[Delhi Univ. B.Com. (Hons.), 1997]
Q 3 – Q1
Solution. We are given : Q1 = 142; and
= 18 ⇒ Q3 = Q1 + 2 × 18 = 142 + 36 = 178
2
If the distribution is symmetrical, then
(Q – Md) – (Md – Q 1 )
Sk (Bowley) = 3
= 0 ⇒ Q3 – Md = Md – Q 1
(Q 3 – Md) + (Md – Q1)
Md = 12 (Q 1 + Q3) = 142 +2 178 = 160
∴
Example 7·16(a). In a frequency distribution the coefficient of skewness based on quartiles is 0·6. If
the sum of the upper and lower quartiles is 100 and median is 38, find the value of upper and lower
quartiles.
(b) The mean, mode an Q.D. of a distribution are 42, 36 and 15 respectively. If its Bowley’s coefficient
of skewness is 1/3, find the values of the two quartiles. Also state the empirical relationship between mean,
mode and median.
[Delhi Univ. B.Com. (Hons.), 2008]
Solution. (a) We are given :
Q + Q1 – 2Md
Sk = 3
= 0·6
…(*) Also
Q 3 + Q1 = 100 and Median = 38
…(**)
Q3 – Q1
Substituting from (**) in (*), we get
100 – 2 × 38
Q3 – Q1 =
0·6
24
Q3 – Q1 = 1000·6– 76 = 0·6
= 40
⇒
Thus, we have : Q3 + Q1 = 100
and
Adding and subtracting, we get respectively
2Q3 = 140
⇒
Q3 = 140
2 = 70;
Q3 – Q1 = 40
and
2Q1 = 60
⇒
Q1 = 60
2 = 30
Hence, the values of the upper and lower quartiles are 70 and 30 respectively.
(b) We are given : Mean = 42, Mode = 36
Q – Q1
and
Q.D. = 3
= 15
⇒
Q3 – Q1 = 30
2
The empirical relation between mean, mode and median is :
Mean – Mode = 3 (Mean – Median)
⇒
Mode = 3 Median – 2 Mean
1
From (1) and (3), we get 36 = 3Md – 2 × 42
⇒
Md = (36 + 84) = 40
3
Bowley’s coefficient of skewness is given by :
Q + Q1 – 2 Md 1
Sk (Bowley) = 3
= (Given)
Q3 – Q1
3
⇒
∴
1
30
Q3 + Q1 = 2 Md + (Q3 – Q1) = 2 × 40 + = 90
3
3
Q3 – Q1 = 30 and Q3 + Q1 = 90.
Adding and subtracting we get respectively.
2 Q3 = 120
⇒
Q3 = 60
and
[From (2) and (4)]
2Q1 = 60
⇒
Q1 = 30.
…(1)
…(2)
…(3)
…(4)
7·16
BUSINESS STATISTICS
Example 7·17. From the information given below, calculate Karl Pearson’s coefficient of skewness
and also quartile coefficient of skewness :
Measure
Mean
Median
Standard deviation
Third quartile
First quartile
Place A
150
142
30
195
62
Place B
140
155
55
260
80
Solution. Place A :
3(M – Md) 3(150 – 142) 8
=
= 10 = 0·8
30
σ
Q + Q1 – 2Md 195 + 62 – 2 × 142 257 – 284
27
Sk (Bowley) = 3
=
=
= – 133 = – 0·203
195 – 62
133
Q3 – Q1
Sk (Karl Pearson) =
Place B :
3(M – Md) 3(140 – 155) 3 × (–15)
9
=
=
= – 11 = – 0·82
55
55
σ
Q + Q1 – 2Md 260 + 80 – 2 × 155 340 – 310 1
Sk (Bowley) = 3
=
=
= 6 = 0·167
260 – 80
180
Q3 – Q1
Note: See Remark 4 to § 7.2.3
EXERCISE 7·2
Sk (Karl Pearson) =
1. What is ‘Skewness’ ? What are the tests of skewness ? Draw different sketches to indicate different types of
skewness and locate roughly the relative positions of mean, median and mode in cash case.
2. Give any three measures of skewness of a frequency distribution. Explain briefly (not exceeding ten lines) with
suitable diagrams the term ‘Skewness’ as mentioned above.
3. Distinguish between Karl Pearson’s and Bowley’s measures of skewness. Which one of these would you prefer
and why ?
[Delhi Univ. MBA, 2000]
4. Explain the meaning of skewness using sketches of frequency curves. State the different measures of skewness
that are commonly used. How does skewness differ from dispersion ?
5. What are quartiles ? How are they used to measure skewness ?
6. Describe Bowley’s measure of skewness. Show that it lies between ±1. Under what conditions these limits are
attained ?
7. The weekly wages earned by one hundred workers of a factory are as follows. Find the absolute measures of
dispersion and skewness based on quartiles and interpret the results.
Weekly wages (’00 Rs.)
No. of workers
Weekly wages (’00 Rs.)
No. of workers
12·5—17·5
12
37·5—42·5
10
17·5—22·5
16
42·5—47·5
6
22·5—27·5
25
47·5—52·5
3
27·5—32·5
14
52·5—57·5
1
32·5—37·5
13
Ans. Q.D. = 7·01; Sk = (Q3 – Md) – (Md – Q1) = 3·34.
8. Find out the coefficient of skewness using Bowley’s formula from the following figures :
Income (in Rs.) : 100—199 200—299 300—399 400—499 500—599 600—699 700—799 800—899
No. of Persons
:
39
25
49
62
38
37
32
18
Ans. Sk = 0·1146.
9. The figures in the adjoining table relate to the size of capital Capital in Lakhs of Rs.
of companies :
1—5
Find out (i) the median size of the capital;
6—10
(ii) the coefficient of skewness with the help of Bowley’s
11—15
measure of skewness.
16—20
What conclusion do you draw from the skewness measured by
21—25
you ?
26—30
31—35
Ans. Median = 23·47 ; Sk (Bowley) = – 0·119.
No. of Companies
———————————————————————————————
——————————————————————————————
20
27
29
38
48
53
70
7·17
SKEWNESS, MOMENTS AND KURTOSIS
10. For the frequency distribution given below, calculate the coefficient of skewness based on the quartiles :
Class Limits
:
10—19
20—29
30—39
40—49
50—59
60—69
70—79
80—89
Frequency
:
5
9
14
20
25
15
8
4
Ans. – 0·103.
11. Calculate the coefficient of skewness based n quartiles and median from the following data :
Variable
:
0—10
10—20
20—30
30—40
40—50
50—60
60—70
70—80
Frequency
:
12
16
26
38
22
15
7
4
[Osmania Univ. B.Com., 1998]
Ans. Q1 = 22·69, Q3 = 45·91, Median (Q2) = 34·21, Sk (Bowley) = 0·008.
12. By using the quartiles, find a measure of skewness for the following distribution.
Annual Sales (Rs. ’000)
Less than
”
”
”
”
”
”
”
”
No. of firms
20
30
40
50
60
Annual Sales (Rs. ’000)
No. of firms
Less than 70
”
” 80
”
” 90
”
” 100
30
225
465
580
634
644
650
665
680
[Bangalore Univ. B.Com., 1998]
Ans. Q1 = 27·18 ; Q3 = 43·90 ; Median = 34·79 ; Sk (Bowley) = 0·0903.
13. Calculate Bowley’s coefficient of skewness for the following data :
Income equal to or more than
No. of Persons
Income equal to or more than
200
150
100
50
0
600
700
800
900
1000
1000
950
700
600
500
100
200
300
400
500
No. of Persons
Assume that income is a continuous variable.
Ans. Sk = – 0·4506.
14. (a) Find out Quartile coefficient of skewness in a series where
Q1 = 18, Q3 = 25, Mode = 21 and Mean = 18.
[Delhi Univ. B.Com. (Pass) 1999]
Hint. Mode = 3Md – 2 Mean
⇒
Md = 19
Ans. 0·714.
(b) Find the coefficient of skewness from the following information :
Difference of two quartiles = 8 ; Mode = 11, Sum of two quartiles = 22 ; Mean = 8.
[Delhi Univ. B.Com. (Hons.), 1997]
Hint. Q3 – Q1 = 8 , Q3 + Q1 = 22
⇒
Q1 = 7, Q3 = 15
;
Mo = 3Md – 2M ⇒
Md =
11 + 16
=9
3
Ans. Sk (Bowley) = 0·5.
15. (a) If the quartile coefficient of skewness is 0·5, quartile deviation is 8 and the first quartile is 16, find the
median of the distribution.
[Delhi Univ. B.A. (Econ. Hons.), 2002]
Ans. Median = 20.
(b) The measure of skewness for a certain distribution is – 0·8. If the lower and the upper quartiles are 44·1 and
56·6 respectively, find the median.
Ans. Md = 55·35.
16. (a) The coefficient of skewness for a certain distribution based on the quartiles is – 0·8. If the sum of the upper
and lower quartiles is 100·7 and median is 55·35, find the values of the upper and lower quartiles.
Ans. Q3 = 56·6; Q1 = 44·1.
7·18
BUSINESS STATISTICS
(b) In a frequency distribution, coefficient of skewness based on quartiles is 0·6. If the sum of upper and lower
quartiles is 100 and median is 38, find the values of lower and upper quartiles.
[Delhi Univ. B.Com. (Hons.), 1993]
Ans. Q1 = 30, Q3 = 70.
17. In a distribution, the difference of the two quartiles is 15 and their sum is 35 and the median is 20. Find the
coefficient of skewness.
Ans. Sk (Bowley) = – 0·33.
18. If the First Quartile is 48 and Quartile Deviation is 6, find the Median (assuming the distribution to be
symmetrical).
[Delhi Univ. B.Com. (Pass), 1996]
Ans. Md = 12 (Q3 + Q1) =
48 + 60
= 54.
2
19. Fill in the blank :
“If Q1 = 6, Q3 = 10 and Bowley’s coefficient of skewness is 0·5, then the value of median will be equal to….”
Ans. Md = 7.
20. Particulars relating to the wage distribution of two manufacturing firms are as follows :
Mean wage
Median wage
Modal wage
Quartiles
S.D.
Firm ‘A’
Rs.
175
172
167
162 ; 178
13
Firm ‘B’
Rs.
180
170
162
165 ; 185
19
Compare the two distributions.
Ans. C.V. (Firm A) = 7·43, C.V. (Firm B) = 10·56; Bowley’s coeff. of skewness (Firm A) = – 0·25, skewness
(Firm B) = 0·5; Karl Pearson coeff. of skewness (Firm A) = 0·615, skewness (Firm B) = 0·947.
21. Find out Bowley’s coefficient of skewness from the following data and show which section is more skewed :
Income in ’00 Rs.
:
55—58
58—61
61—64
64—67
67—70
Section A
:
12
17
23
18
11
Section B
:
20
22
25
13
4
Ans. Bowley’s coefficient of skewness (A) = – 0·0061. ; Bowley’s coefficient of skewness (B) = – 0·06.
For the comparison of these negative skewnesses, we compare their absolute values. Section B is more
skewed.
7·3. MOMENTS
Moment is a term generally used in physics or mechanics and provides us
a measure of the turning or the rotating effect of a force about some point. The
moment of a force, say, F about some point P is given by the product of the
magnitude of the force (F) and the perpendicular distance (p) between the point
of reference and direction of the force [Fig. 7·4] i.e.,
Moment = p × F
However, the term moment as used in physics has nothing to do with the
moment used in Statistics, the only analogy being that in Statistics we talk of
moment of random variable about some point and these moments are used to
describe the various characteristics of a frequency distribution viz., central
tendency, dispersion, skewness and kurtosis.
Let the random variable X have a frequency distribution
X | x 1 x2
x3 … … xn |
———————————————————————————————————————————————
Let
f | f1 f3
f3 … … fn | ∑f = N
–x = ∑ fx = ∑ fx , be its arithmetic mean.
∑f
N
p
p
F
Fig. 7·4.
7·19
SKEWNESS, MOMENTS AND KURTOSIS
7·3·1. Moments about Mean. The rth moment of X about the mean –x , usually denoted by μr [where μ
is the letter Mu of the Greek alphabet] is defined as
μr = 1N ∑ f (x – –x )r ; r = 0, 1, 2, 3,…
…(7·19)
[
1
=N
f1 (x1 – –x ) r + f2 (x2 – –x ) r + … + fn (xn – –x ) r
In particular putting r = 0 in (7·19), we get
1
μ0 = N ∑ f (x – –x ) 0 = N1 ∑ f = 1N · N
]
…(7·19a)
[·. · (x – –x ) 0 = 1 and ∑f = N]
μ0 = 1.
…(7·20)
Putting r = 1 in (7·19), we get
μ1 = N1 ∑
f (x – –x ) = 0,
…(7·21)
because the algebraic sum of deviations of a given set of observations from their mean is zero. Thus, the
first moment about mean is always zero.
Again taking r = 2, we get
μ2 = N1 ∑ f (x – –x)2 = σx2
…(7·22)
Hence, the second moment about mean gives the variance of the distribution.
Also
μ = 1 ∑ f (x – –x)3
and
μ = 1 ∑ f (x – –x)4
3
4
N
N
…(7·23)
7·3·2. Moments about Arbitrary Point A. The rth moment of X about any arbitrary point A, usually
denoted by μr′ is defined as :
1
μr′ = N
∑ f (x – A)r
1
;
r = 0, 1, 2, 3,…
[
…(7·24)
]
= N f1 (x1 – A)r + f2(x2 – A)r + … + fn (xn – A)r
…(7·24a)
In particular taking r = 0 and r = 1 in (7·24), we get respectively
1
1
μ0 ′ = N ∑ f (x – A)0 = N ∑ f = 1
1
…(7·25)
1
μ1 ′ = N ∑ f(x – A) = N [∑ fx – A∑ f]
1
[
]
(·. · A is constant)
1
= N ∑fx – AN = N ∑ fx – A
= –x – A
…(7·26)
–x = A + μ′
⇒
1
where μ 1 ′ is the first moment about the point ‘A’.
Taking r = 2, 3, 4 in (7·24), we get respectively
1
μ2 ′ = Second moment about A = N ∑ f (x – A)2
μ3 ′ =
μ4 ′ =
1
Third moment about A = N ∑ f (x – A)3
1
Fourth moment about A = N ∑ f (x – A)4
…(7·27)
…(7·28)
…(7·28 a)
…(7·28 b)
Remarks 1. μr, the rth moment about the mean ; r = 1, 2, 3, … are also called the central moments and
μ′r, the rth moment about any arbitrary point A are also known as raw moments.
2. In particular, if we take A = 0 in (7·27), we get
–x = 0 + μ ′ (about origin)
…(7·28 c)
1
Hence, the first moment about origin gives mean.
7·3·3. Relation between Moments about Mean and Moments about Arbitrary Point ‘A’.
We have
μr = μr′ – rC 1 μ′r – 1μ1 ′ + rC 2 μ′r – 2 μ′1 2 – rC 3 μ′r – 3 μ′1 3 + … + (–1)r μ1′r
…(7·29)
7·20
BUSINESS STATISTICS
Remarks 1. We summarise below the important results on moments :
μ0 = 1 and μ1 = 0
Mean (x–) = A + μ ′
}
1
′2
Variance = σx = μ2 = μ2′ – μ1
μ3 = μ3 ′ – 3μ2 ′ μ1 ′ + 2μ1 ′3
μ4 = μ4 ′– 4μ3 ′ μ1 ′ + 6μ2 ′ μ1 ′2 – 3μ1 ′4)
2
…(7·30)
}
…(7·31)
The results in (7·30) and (7·31) are of fundamental importance in Statistics and should be committed to
memory. Thus, if we know the first four moments about any arbitrary point A, we can obtain the measures
of central tendency [–x = μ1 ′ (about origin)], dispersion (μ2 = σ 2 ), skewness (μ3 or β1) and kurtosis (β 2 ).
[The last two measures are discussed in the Sections § 7·5 and § 7·6]. Since, these four measures enable us
to have a fairly good idea about the nature and the form of the given frequency distribution, in practice we
generally compute only the first four moments and not the higher moments.
2. In case of a symmetrical distribution, if the deviations of the given observations from their
arithmetic mean are raised to any odd power, the sum of positive deviations equals the sum of the negative
deviations and accordingly the overall sum is zero, i.e., if the distribution is symmetrical then :
∑ f (x – –x ) = ∑ f (x – –x ) 3 = ∑ f (x – –x ) 5 = ∑ f (x – –x ) 7 = … = 0
Dividing by N = ∑ f, we get for a symmetrical distribution :
μ 1 = μ 3 = μ5 = μ7 = … = 0
⇒
μ2 r + 1 = 0 ; r = 0, 1, 2, 3,…
…(7·32)
Hence, for a symmetrical distribution, all the odd order moments about mean vanish. Accordingly, odd
order moments, specially 3rd moment is used as a measure of skewness.
3. We have
μ2 = μ2 ′ – μ1 ′2
Since μ1′2, being the square of a real quantity is always non-negative, we get
μ2 = μ2 ′ – (some non-negative quantity)
⇒
μ 2 ≤ μ2 ′
∴
Variance ≤ Mean square deviation or S.D. ≤ Root mean square deviation …(7·33)
4. For obtaining the moments of a grouped (continuous) frequency distribution, if we change the scale
also in X by taking
x–A
d =
⇒
x – A = hd
…(7·33a)
h
then
μr′ = rth moment of X about point X = A
1
1
= N ∑ f (x – A)r = N ∑ f . hr dr
⇒
μr′ =
1
hr . N ∑
[From (7·33a)]
f dr
…(7·33b)
In particular, we have
1
μ1 ′ = h . N ∑ f d;
1
μ 2 ′ = h2 . N ∑ f d 2 ;
1
μ 3 ′ = h3 .N ∑ f d 3 ;
1
μ 4 ′ = h4 .N ∑ f d 4
…(7·33c)
Finally, on using the relation (7·31), we obtain the moments about mean.
For numerical computations, if the mean of the distribution comes out to be integral (i.e., a whole
number), then it is convenient to obtain the moments about mean directly by the formula
1
μr = N
∑ f (x – –x ) r
…(7·33d)
However, for grouped (continuous) frequency distribution, the calculations are simplified by changing
the scale also in X. If we take
x – –x
z =
⇒
x – –x = hz
…(7·34)
h
7·21
SKEWNESS, MOMENTS AND KURTOSIS
then, we have
1
μr = N ∑ f . (hz)r
⇒
μ r = hr .
[
1
N∑
f zr
]
…(7·35)
a formula which is more convenient to use.
4. Converse. We can obtain μr′ in terms of μ r as given below :
μ′r = μr + rC 1 μr – 1 μ1′ + rC 2 μr – 2 μ1′2 + … + μ1 ′r,
…(7·36)
In particular, taking r = 2, 3 and 4 in (7·36) and simplifying we shall get respectively :
μ 2 ′ = μ 2 + μ 1 ′2
μ3 ′ = μ3 + 3μ2 μ1′ + μ1 ′3
…(7·37)
μ4 ′ = μ4 + 4μ3 μ1′ + 6μ2 μ1′2 + μ1′4
These formulae enable us to compute moments about any arbitrary point, if we are given mean and
moments about the mean.
}
7·3·4. Effect of Change of Origin and Scale on Moments about Mean. Let us change the origin and
scale in the variable x to obtain a new variable u as defined below :
x–a
u =
⇒
x = a + hu
…(7·38)
h
Then,
μr (x) = hrμr (u)
…(7·39)
Hence, rth moment of the variable x about its mean is equal to hr times the rth moment of the variable
u about its mean. The result does not depend on ‘a’. Hence, we conclude that the moments about mean (i.e.,
central moments) are invariant under change of origin but not of scale.
7·3·5. Sheppard’s Correction for Moments. In case of grouped or continuous frequency distribution,
for the calculation of moments, the value of the variable X is taken as the mid-point of the corresponding
class. This is based on the assumption that the frequencies are concentrated at the mid-points of the
corresonding classes. This assumption is approximately true for distributions which are symmetrical or
moderately skewed and for which the class intervals are not greater than one-twentieth (1/20th) of the range
of the distribution. However, in practice, this assumption is not true in general and consequently some
error, known as ‘grouping error’ is introduced in the calculation of moments. W.F. Sheppard proved that if
(i) the frequency distribution is continuous, and
(ii) the frequency tapers off to zero in both directions,
the effect due to grouping at the mid-point of the intervals can be corrected by the following formulae,
known as Sheppard’s corrections :
h2
⎫
⎪
⎪
⎬
⎪
⎪⎭
μ2 (corrected) = μ2 – 12
μ3 (corrected) = μ3
μ4 (corrected) =
μ4 – 12 h2 μ2
7 4
+ 240
h
…(7·40)
where h is the width of the class interval.
Remark. This correction is valid only for symmetrical or slightly asymmetrical continuous
distributions and cannot be applied in the case of extremely asymmetrical (skewed) distributions like
J-shaped or inverted J-shaped or U-shaped distributions. As a safeguard against sampling errors, this
correction should be applied only if the total frequency is fairly large, say, greater than 1000.
7·3·6. Charlier Checks. We have, on using binomial expansion for positive integral index,
x+1 =x+1
(x + 1)2 = x 2 + 2x + 1
(x + 1)3 = x 3 + 3x2 + 3x + 1
(x + 1)4 = x 4 + 4x3 + 6x2 + 4x + 1
}
…(*)
7·22
BUSINESS STATISTICS
Multiplying both sides of (*) by f and adding over different values of the variable X, we get the
following identities,
∑ f (x + 1) = ∑ f x + N
∑ f (x + 1)2 = ∑ f x2 + 2∑ f x + N
…(7·41)
∑ f (x + 1)3 = ∑ f x3 + 3∑ f x2 + 3∑ f x + N
∑ f (x + 1)4 = ∑ f x4 + 4∑ f x3 + 6∑ f x2 + 4 ∑ f x + N
}
These identities are known as Charlier checks and are used in checking the calculations in the
computation of the first four moments.
β ) AND GAMMA (γγ ) COEFFICIENTS BASED ON
7·4. KARL PEARSON’S BETA (β
MOMENTS
Prof. Karl Pearson defined the following four coefficients based on the 1st four central moments.
β1 =
μ3 2
μ2 3
…(7·42)
β2 =
μ4 μ4
=
μ 2 2 σ4
…(7·43)
μ3
μ
= 3
μ2 3/2 σ3
γ1 = ⎯
√⎯β1 =
γ2 = β 2 – 3 =
(·.· μ2 = σ2 )
…(7·44)
μ4
–3
μ2 2
…(7·45)
It may be stated here that these coefficients are pure numbers independent of units of measurement and
as such can be conveniently used for comparative studies. In practice they are used as measures of
skewness and kurtosis as discussed in the following sections.
Remark. Sometimes, another coefficient based on moments viz., Alpha ( α) coefficient is used. Alpha
coefficients are defined as
μ
μ
μ
μ
α1 = 1 = 0, α2 = 22 = 1, α3 = 33 = √
β1 = γ1, α4 = 44 = β2
…(7·46)
⎯
⎯
σ
σ
σ
σ
7·5. COEFFICIENT OF SKEWNESS BASED ON MOMENTS
Based on the first four moments, Karl Pearson’s coefficient of skewness becomes
Sk =
√⎯β1 (β2 + 3)
⎯
…(7·47)
2(5β2 – 6β1 – 9)
where β 1 and β2 are Pearson’s coefficients defined in (7·42) and (7·43) in terms of the first four central
moments. Formula (7·47) will give positive skewness if M > Mo and negative skewness if M < Mo. Sk = 0,
if β1 = 0 or β2 + 3 = 0
⇒ β 2 = –3.
1
1
But
μ = ∑ f (x – –x )4 > 0 and μ = ∑ f (x – –x ) 2 > 0
4
∴
β2 =
N
2
N
μ4
> 0
μ2 2
Since β2 cannot be negative, Sk = 0 if β 1 = 0 or if μ3 = 0. Hence, for a symmetrical distribution, β1 = 0.
Accordingly, β1 may be taken as a relative measure of skewness based on moments.
Remark. The coefficient β1 as a measure of skewness has a serious limitation. μ3 being the sum of the
cubes of the deviations from the mean may be positive or negative but μ3 2 is always positive. Also μ2 being
the variance is always positive. Hence, β1 = μ3 2 /μ 2 3 , is always positive. Thus β1 , as a measure of skewness
7·23
SKEWNESS, MOMENTS AND KURTOSIS
is not able to tell us about the direction (positive or negative) of skewness. This drawback is removed in
Karl Pearson’s coefficient [Gamma One, (γ1)] which is defined as the positive square root of β1 , i.e. ;
μ3
μ
γ1 = + ⎯
= 33
…(7·48)
√⎯β1 = μ 3/2
σ
2
Thus, the sign of skewness depends upon μ3 . If μ3 is positive we get positive skewness and if μ3 is
negative, we get negative skewness.
7·6. KURTOSIS
So far we have studied three measures viz., central tendency, dispersion and skewness to describe
the characteristics of a frequency distribution. However, even if we know all these three measures we are
not in a position to characterise a distribution completely. The following diagram (Fig. 7.5) will clarify the
point.
A
B
A – Lepto-kurtic
B – Meso-kurtic
C – Platy-kurtic
C
Mean
Fig. 7·5.
All the three curves are symmetrical about the mean and have same variation (range). In order to
identify a distribution completely we need one more measure which Prof. Karl Pearson called ‘convexity of
the curve’ or its ‘Kurtosis’. While skewness helps us in identifying the right or left tails of the frequency
curve, kurtosis enables us to have an idea about the shape and nature of the hump (middle part) of a
frequency distribution. In other words, kurtosis is concerned with the flatness or peakedness of the
frequency curve.
Curve of type B which is neither flat nor peaked is known as Normal curve and shape of its hump is
accepted as a standard one. Curves with humps of the form of normal curve are said to have normal
kurtosis and are termed as meso-kurtic. The curves of the type A., which are more peaked than the normal
curve are known as lepto-kurtic and are said to lack kurtosis or to have negative kurtosis. On the other
hand, curves of the type C, which are flatter than the normal curve are called platy-kurtic and they are said
to possess kurtosis in excess or have positive kurtosis.
As a measure of kurtosis, Karl Pearson gave the coefficient Beta two (β2 ) or its derivative Gamma two
(γ2) defined as follows :
β2 =
μ4
μ
= 44
2
μ2
σ
…(7·49)
μ4
μ – 3σ4
–3 = 4 4
…(7·50)
4
σ
σ
For a normal or meso-kurtic curve (Type B), β2 = 3 or γ2 = 0. For a lepto-kurtic curve (Type A),
β2 > 3 or γ2 > 0 and for a platy-kurtic curve (Type C), β2 < 3 or γ2 < 0.
γ2 = β2 – 3 =
It is interesting to quote here the words of a British statistician W.S. Gosset (who wrote under the pen
name of Student), who very humorously explains the use of the terms platy-kurtic and lepto-kurtic in the
following sentence : “Platykurtic curves like the platypus, are squat with short tails ; lepto-kurtic curves
are high with long tails like the kangaroos noted for leaping.”
7·24
BUSINESS STATISTICS
Gosset’s little but humorous sketch is given below :
Platypus
Kangaroos
Fig. 7.6 (a).
Fig. 7.6 (b).
Remarks 1. The Pearsonian coefficients β1 and β2 are independent of the change of origin and scale
i.e., if we take :
X – A,
U=
h>0
then
β1(X) = β 1 (U)
and
β2(X) = β 2 (U)
h
2. For a discrete distribution β2 ≥ 1.
In fact, a much stronger result holds. We have : β 2 ≥ β1 + 1.
Example 7·18. “For a symmetrical distribution, all central moments of odd order are zero.” Comment.
Solution. The statement is always true i.e., for a symmetrical distribution,
μ 1 = μ 3 = μ5 = … = 0
i.e.,
μ2n + 1 = 0 ; (n = 0, 1, 2,…)
[For detailed discussion see Remark 2, § 7·3·3]
Example 7·19. Calculate the first four moments about the mean for the following data and comment on
the nature of the distribution :
x:
f:
1
1
2
6
3
13
4
25
5
30
6
22
7
9
8
5
9
2
[Delhi Univ. B.Com. (Hons.), 1999]
Solution.
CALCULATIONS FOR MOMENTS
x
f
1
2
3
4
5
6
7
8
9
1
6
13
25
30
22
9
5
2
∑ f = 113
d=x–5
–4
–3
–2
–1
0
1
2
3
4
∑d=0
fd
–4
–18
–26
–25
0
22
18
15
8
∑ fd = –10
Moments About the Point x = 5
d = x – A = x – 5, (A = 5) ; N = ∑ f = 113
∑ fd –10
μ1 ′ =
= 113 = – 0·0885
N
∑ fd2 282
μ2 ′ =
= 113 = 2·4956
N
∴
Mean = A + μ1′ = 5 + (– 0·0885) = 4·9115
fd2
fd3
fd4
16
54
52
25
0
22
36
45
32
– 64
–162
–104
–25
0
22
72
135
128
256
486
208
25
0
22
144
405
512
∑ fd2 = 282
∑ fd3 = 2
∑ fd4 = 2058
∑ fd 3 2
= 113 = 0·0177
N
∑ fd 4 2058
μ4 ′ =
= 113 = 18·2124
N
μ3 ′ =
7·25
SKEWNESS, MOMENTS AND KURTOSIS
Moments About Mean
μ1 = 0
μ4 = μ4 ′ – 4 μ3 ′ μ1 ′ + 6 μ2 ′ μ1 ′2 – 3 μ1′4
2
2
μ2 = μ2 ′ – μ1 ′ = 2·4956 – (– 0·0885)
= 18·2124 – 4 × (0·0177) × (–0·0885)
= 2·4956 – 0·0078 = 2·4878
+ 6 × 2·4956 × (– 0·0885)2 – 3 (– 0·0885)4
3
μ3 = μ3 ′ – 3μ2 ′ μ1 ′ + 2μ1 ′
= 18·2124 + 0·00626 + 0·11728 – 0·000184
= 0·0177 – 3 × 2·4956 × (– 0·0885)
= 18·3357
+ 2 × (– 0·0885)3
= 0·0177 + 0·66258 – 0·001386
= 0·6789
Moment Coefficients of Skewness
μ 2 (0·6789)2
0·4609
β1 = 3 3 = (2·4878)
3 =
15·3974 = 0·0299
μ2
μ3
μ3
0·6789
γ1 = 3/2
=
= 3·9239 = 0·173 or γ1 = ⎯
0·0299 = + 0·173 (·.· μ3 is positive)
√⎯β1 = + ⎯√⎯⎯⎯⎯
μ2
μ2⎯
μ
√⎯ 2
Kurtosis
β2 =
μ4 18·3357
=
= 18·3357
6·1891 = 2·9626
μ2 2 (2·4878)2
⇒
γ2 = β2 – 3 = – 0·0373.
Interpretation
Since γ1 = 0·173, the given distribution is very slightly positively skewed.
Since β2 = 2·9626 ~– 3, the given distribution is more or less normal or meso-kurtic. However, since
β2 < 3 (γ2 = – 0·04), the given frequency curve is slightly Platy-kurtic and is slightly flatter than the normal
curve.
Example 7·20. From the following data, calculate moments about assumed mean 25 and convert them
into central moments :
X
:
0—10
10—20
20—30
30—40
f
:
1
3
4
2
[Delhi Univ. B.Com. (Hons.), 2000]
Solution.
CALCULATIONS FOR MOMENTS ABOUT THE POINT X = 25
Class
Mid-value
(X)
f
d = 10
fd
fd2
fd3
fd4
0—10
10—20
20—30
30—40
5
15
25
35
1
3
4
2
–2
–1
0
1
–2
–3
0
2
4
3
0
2
–8
–3
0
2
16
3
0
2
Total
N = ∑ f = 10
Moments about the Point X = 25
∑ fd 10 × (–3)
μ1 ′ = h .
= 10 = –3
N
∑ fd2 102 × 9
μ2 ′ = h2 .
= 10 = 90
N
x – 25
∑ fd = –3
∑ fd2 = 9
∑ fd3 = – 9
∑ fd4 = 21
h3 × ∑ fd3 103 × (–9)
= 10 = –900
N
h4 . ∑ fd4 104 × 21
μ4 ′ =
= 10 = 21,000
N
μ3 ′ =
Central Moments (Moments About Mean)
μ1 = 0
μ2 = μ2 ′ – μ1 ′2 = 90 – 9 = 81
μ3 = μ3 ′ – 3μ2 ′μ1 ′ + 2μ1 ′3
μ4 = μ4 ′ – 4μ3 ′μ1 ′ + 6μ2 ′ μ1 ′2 – 3μ1 ′4
3
= –900 – 3 × 90 × (–3) + 2 (–3)
= 21,000 – 4 × (–900) × (–3) + 6 × 90 × (–3)2 – 3 × (–3)4
= –900 + 810 – 54 = –144
= 21,000 – 10,800 + 4,860 – 243
= 14,817
7·26
BUSINESS STATISTICS
Example 7·21. The first three moments of a distribution about the value 67 of the variable are 0·45,
8·73 and 8·91. Calculate the second and third central moments, and the moment coefficient of skewness.
Indicate the nature of the distribution.
[Delhi Univ. B.A. (Econ. Hons. I), 2000]
Solution. In the usual notations we are given :
A = 67, μ 1 ′ = 0·45, μ2 ′ = 8·73 and μ3 ′ = 8·91
The second and third central moments are given by :
μ2 = μ2 ′ – μ1 ′2 = 8·73 – (0·45)2 = 8·73 – 0·2025 = 8·5275
μ3 = μ3 ′ – 3 μ2 ′ μ1 ′ + 2μ1 ′3 = 8·91 – 3 × 8·73 × 0·45 + 2 × (0·45)3
= 8·91 – 11·7855 + 0·18225 = –2·6933
Hence, the variance of the distribution is
σ2 = μ2 = 8·5275
⇒
σ(s.d.) = ⎯√⎯⎯⎯⎯
8·5275 = 2·9202
Since μ3 is negative, the given distribution is negatively skewed. In other words, the frequency curve
has a longer tail towards the left. Karl Pearson’s moment coefficient of skewness is given by :
μ3
μ3
μ2
–2·6933
γ1 = 3/2
=
= 8·5275
= –2·6933
= –0·1082 ⇒ β1 = 3 3 = (– 0·1082)2 = 0·0117
×
2·9202
24·9020
μ2
μ2
μ2⎯
√⎯μ2
Since γ1 and β1 are approximately zero, the given distribution is approximately symmetrical.
Example 7·22. The first four moments of a distribution about the origin are 1, 4, 10 and 46
respectively. Obtain the various characteristics of the distribution on the basis of the information given.
Comment upon the nature of the distribution.
Solution. We are given the first four moments about origin. In the usual notations we have :
A = 0, μ 1 ′ = 1, μ2 ′ = 4, μ3 ′ = 10
and
μ4 ′ = 46
The measure of central tendency is given by :
Mean ( –x ) = First moment about origin = μ1′ = 1
The measure of dispersion is given by :
Variance (σ2 ) = μ2 = μ2′ – μ1 ′2 = 4 – 1 = 3
⇒
s.d. (σ) = ⎯
√ 3 = 1·732
′3
μ3 = μ3 ′ – 3μ2 ′ μ1 ′ + 2μ1 = 10 – 3 × 4 × 1 + 2 × 1 = 10 – 12 + 2 = 0
Karl Pearson’s moment coefficient of skewness is given by :
γ1 =
μ3
=0
μ2 3/2
⇒
μ 2
β1 = μ 33 = 0
2
Since γ1 = 0, the given distribution is symmetrical, i.e., Mean = Median = Mode,
for the given distribution. Moreover, the quartiles are equidistant from the median i.e.,
Q3 – Median = Median – Q1
μ4 = μ4 ′ – 4μ3 ′ μ1 ′ + 6μ2 ′ μ1 ′2 – 3μ1 ′4
= 46 – 4 × 10 × 1 + 6 × 4 × 1 – 3 × 1 = 46 – 40 + 24 – 3 = 27
Hence, Karl Pearson’s measure of Kurtosis is given by :
β2 =
μ4 27
=
=3
μ 2 2 32
⇒
Since β2 = 3, the given distribution is Normal (Meso-kurtic).
γ2 = β2 – 3 = 0
7·27
SKEWNESS, MOMENTS AND KURTOSIS
Since β1 = 0 and β2 = 3, the given distribution is a normal distribution with
Mean ( –x ) = 1 and s.d. (σ) = √
⎯ 3 = 1·732.
Example. 7.23. Given that the mean of a distribution is 5, variance is 9 and the moment coefficient of
skewness is – 1, find the first three moments about origin.
[Delhi Univ. B.A. (Econ. Hons.), 2009]
—
Solution. We are given Mean (X ) = 5, Variance (σ2 ) = μ2 = 9
γ1 =
and
⇒
σ = 3;
μ3
= – 1 ⇒ μ3 = – 27.
σ3
We want moments about the point A = 0.
—
Mean ( X ) = A + μ1′
μ2 = μ2′ – μ1
′2
μ3 = μ3′ – 3μ2 ′μ1′ + 2μ1
′3
⇒
5 = 0 + μ1 ′
⇒
⇒
9 = μ2 ′ –
⇒
⇒
52
– 27 = μ3 ′ – 3 × 34 × 5 + 2 ×
μ1 ′ = 5
μ2 ′ = 34
53
⇒ μ3 ′ = – 27 + 510 – 250 = 233
Example 7·24. If √
⎯⎯β1 = 1, β 2 = 4 and variance = 9, find the values of third and fourth central
moments and comment upon the nature of the distribution.
[Delhi Univ. B.A. (Econ. Hons.), 2007]
Solution. We are given ⎯
√⎯β1 = +1, β2 = 4 and Variance = μ 2 = 9
Also
√⎯β1 = 1
⎯
⇒
μ3 2
μ2 3
β2 = 4
⇒
μ4
=4
μ2 2
=1
…(*)
⇒
μ3 = μ23/2 . 1 = 93/2 . 1 = 33 = 27
⇒
μ4 = 4 × 9 × 9 = 324
[
or β1 =
μ3
σ3
]
∴ μ3 = 27 and μ4 = 324.
Nature of the Distribution. Since γ1 = √
⎯⎯β1 ≠ 0 and γ1 = 1, the distribution is moderately positively
skewed. (Q μ3 > 0).
Also β2 = 4 > 3. Hence, the given distribution is lepto-kurtic i.e., more peaked than the normal curve.
Example 7·25. For a meso-kurtic distribution, the first moment about 7 is 23 and the second moment
about origin is 1000. Find the coefficient of variation and the fourth moment about the mean.
[Delhi Univ. B.Com. (Hons.), 2008; Delhi Univ. B.A. (Econ. Hons.), 2002]
Solution. Since the distribution is given to be meso-kurtic, we have :
β2 = 3
μ4
=3
μ2 2
⇒
First moment about ‘7’ is 23
∴
⇒
μ4 = 3μ22
…(i)
μ1′ (about 7) = 23 (Given)
i.e.,
Mean = 7 + μ1 ′ = 7 + 23 = 30
…(ii)
But mean is the first moment about origin.
∴
μ1 ′ (about origin) = 30
Moments About Origin
μ1 ′ = Mean = 30; μ 2 ′ = 1,000 (Given)
∴
μ2 = μ2′ – μ1 ′2 = 1000 – 30 2 = 100
⇒
Variance (σ2 ) = 100
× s.d. 100 × 10
Coefficient of Variation (C.V.) = 100
= 33·33
Mean = 30
⇒
s.d. (σ) = 10
7·28
BUSINESS STATISTICS
Substituting the value of μ2 = 100, in (i), the fourth moment about mean is given by :
μ4 = 3 × 1002 = 30,000.
Example 7·26. (a) You are given the following information :
Class
f
Class
f
Class
f
40—41
1
45—46
20
50—51
19
41—
2
46—
38
51—
14
42—
4
47—
52
52—
6
43—
7
48—
40
53—
4
49—50
29
54—55
2
44—45
12
Calculate the first four moments about 47·5. Convert these into moments about the mean and calculate
β1 and β2.
(b) Also apply Sheppard’s corrections to moments.
Solution.
CALCULATIONS FOR MOMENTS
Class
Mid-value
(X)
40—41
41—42
42—43
43—44
44—45
45—46
46—47
47—48
48—49
49—50
50—51
51—52
52—53
53—54
54—55
40·5
41·5
42·5
43·5
44·5
45·5
46·5
47·5
48·5
49·5
50·5
51·5
52·5
53·5
54·5
Total
f
1
2
4
7
12
20
38
52
40
29
19
14
6
4
2
d = X – 47·5
fd
–7
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
6
7
–7
–12
–20
–28
–36
– 40
–38
0
40
58
57
56
30
24
14
N = 250
fd2
49
72
100
112
108
80
38
0
40
116
171
224
150
144
98
∑ fd = 98
fd3
fd4
–343
– 432
–500
– 448
–324
–160
–38
0
40
232
513
896
750
864
986
2391
2592
2500
1792
972
320
38
0
40
464
1539
3584
3750
5184
4802
∑ fd2 = 1502 ∑ fd3 = 1736
First four Moments about A = 47·5. Since h = 1, we have :
98
μ1 ′ = 1N ∑ fd = 250 = 0·392
μ2 ′ =
1
N∑
fd2 = 1502
250 = 6·008
μ3 ′ =
μ4′ =
1
1736
3
N ∑ fd = 250 = 6·944
1
29968
4
N ∑fd = 250 = 119·872
First four Moments about Mean :
Mean = 47·5 + μ1 ′ = 47·5 + 0·392 = 47·892
μ1 = 0
μ2 = μ2 ′ – μ1 ′2 = 6·008 – (0·392)2 = 6·008 – 0·1537 = 5·854
μ3 = μ3 ′ – 3μ2 ′ μ1 ′+ 2μ1 ′3 = 6·944 – 3 × 6·008 × 0·392 + 2 × (0·392)3
⇒
= 6·944 – 18·024 × 0·392 + 2 × 0·0602
μ3 = 6·944 – 7·0654 + 0·1204 = – 0·001 ~– 0
∑ fd4 = 29968
7·29
SKEWNESS, MOMENTS AND KURTOSIS
μ4 = μ4 ′ – 4μ3 ′ μ1 ′ + 6μ2 ′ μ1 ′2 – 3μ1 ′4
= 119·872 – 4 × 6·944 × 0·392 + 6 × 6·008 × (0·392)2 – 3 × (0·392)4
= 119·872 – 10·888 + 5·539 – 0·071 = 114·452
μ 2
μ
114·452
114·452
β1 = 3 3 ~– 0 ;
(·. · μ3 = 0)
and
β2 = 42 = (5·854)
2 =
34·269 = 3·34
μ2
μ2
(b) Sheppard’s Corrections. Using (7·40), we have :
μ2 (Corrected) = μ2
h2
– 12 =
1
5·854 – 12
(·. · h = Magnitude of class interval = 1)
= 5·854 – 3·083 = 5·771
μ3 (Corrected) = μ3 ~– 0
7 4
μ4 (Corrected) = μ4 – 12 h2 μ2 + 240
h = 114·452 – 12 × 5·854 +
7
240
(·. · h = 1)
= 114·452 – 2·927 + 0·029 = 111·554.
Example 7.27. For a distribution, it has been found that the first four moments about 27 are 0, 256,
– 2871 and 188462 respectively. Obtain the first four moments about zero. Also calculate the value of β1
and β2, and comment.
[Delhi Univ. B.Com. (Hons.) 2007, 2006]
Solution. We are given :
A = 27;
μ1′ = 0,
μ2 ′ = 256,
μ3′ = – 2871;
μ4 ′ = 188462
Mean = A + μ1 ′ = 27 + 0 = 27
Hence, the moments about ‘A = 27’ are same as moments about mean.
⇒
μ2 = μ2′ = 256,
μ3 = μ3′ = – 2871;
μ4 = μ4′ = 188462
Moments about Origin
μ1 ′ = Mean = 27
μ2 ′ = μ2 + μ1′2 = 256 + 272 = 256 + 729 = 985
μ3 ′ = μ3 + 3μ2 μ1′ + μ1 ′3 = – 2871 + 3 × 256 × 27 + 273
= – 2871 + 20736 + 19683 = 37548
μ4 ′ = μ4 + 4μ3 μ1′ + 6μ2 μ1′2 + μ1′4
β1
= 188462 + 4 × (– 2871) × 27 + 6 × 256 × 272 + 27 4
= 188462 – 310068 + 1119744 + 531441 = 1529579
μ 2 (– 2871)2 8242641
μ3
– 2871
– 2871
= 33 =
=
= 0·4913 ; γ1 = 3/2
=
=
= – 0·701
μ2
(256) 3
16777216
μ2
⎯⎯⎯⎯⎯⎯⎯
√
16777216 4096
β2 =
μ4 188462 188462
=
=
= 2·876
μ2 2 (256) 2
65536
⇒
γ2 = β2 – 3 = – 0·124
Since β1 ≠ 0, the distribution is not symmetrical.
Since μ3 < 0, γ1 = – 0·701 < 0, the distribution is moderately negatively skewed.
β2 < 3
⇒
Distribution is moderately platy-kurtic.
Example 7·28. The first four moments of a distribution about the value 4 of the variable are –1·5, 17,
–30 and 108. Find the moments about mean, β1 and β2.
Find also the moments about (i) the origin, and (ii) the point x = 2.
Solution. In the usual notations, we are given A = 4 and μ1′ = –1·5, μ2 ′ = 17, μ3 ′ = –30 and μ4 ′ = 108.
7·30
BUSINESS STATISTICS
Moments about Mean :
μ2 = μ2 ′ – μ1 ′2 = 17 – (–1·5)2 = 17 – 2·25
μ4 = μ4 ′ – 4μ3 ′ μ1 ′ + 6μ2 ′ μ1 ′2 – 3μ1 ′4
= 14·75
= 108 – 4(–30) (–1·5) + 6(17) (–1·5)2 – 3(–1·5)4
μ3 = μ3 ′ – 3μ2 ′ μ1 ′ + 2μ1 ′3
= 108 – 180 + 229·5 – 15·1875
3
= 142·3125
= –30 – 3 × (17) × (–1·5) + 2(–1·5)
= –30 + 76·5 – 6·75 = 39·75
μ 2 (39·75)2
μ
142·3125
Hence,
β1 = 3 3 = (14·75)3 = 1580·06
;
β2 = 42 = 142·3125
3209·05 = 0·4924
(14·75)2 = 217·5625 = 0·6541
μ2
μ2
Also
Mean ( –x ) = A + μ ′ = 4 + (–1·5) = 2·5
1
Note. Since for a discrete distribution, β2 ≥ 1, [See Remark 2 to § 7.6], there seems to be some error in
the problem.
Moments about Origin. We have
–x = 2·5,
μ = 14·75,
μ = 39·75
and
μ = 142·31 (approx.)
2
3
4
We know –x = A + μ1 ′, where μ 1 ′ is the first moment about the point x = A. Taking A = 0, we get the
first moment about origin as
μ1 ′ = mean = 2·5.
Using (7·35), we get
μ2 ′ = μ2 + μ1′2 = 14·75 + (2·5)2 = 14·75 + 6·25 = 21
μ3 ′ = μ3 + 3μ2 μ1 ′ + μ1 ′3 = 39·75 + 3(14·75)(2·5) + (2·5)3
= 39·75 + 110·625 + 15·625 = 166
μ4 ′ = μ4 + 4μ3μ1 ′ + 6μ2 μ1 ′2 + μ1′4
= 142·3125 + 4(39·75)(2·5) + 6(14·75)(2·5) 2 + (2·5)4
= 142·3125 + 397·5 + 553·125 + 39·0625 = 1132
Moments about the Point x = 2. We have –x = A + μ ′. Taking A = 2, the first moment about the point
1
x = 2 is
μ1 ′ = –x – 2 = 2·5 – 2 = 0·5
Using (7·35), we get
μ2 ′ = μ2 + μ1′2 = 14·75 + 0·25 = 15
μ3 ′ = μ3 + 3μ2μ1 ′ + μ1 3 = 39·75 + 3(14·75)(0·5) + (0·5)3
= 39·75 + 22·125 + 0·125 = 62
μ4 ′ = μ4 + 4μ3μ1 ′ + 6μ2 μ1 ′2 + μ1′4
= 142·3125 + 4(39·75)(0·5) + 6(14·75)(0·5) 2 + (0·5)4
= 142·3125 + 79·5 + 22·125 + 0·0625 = 244
Example 7·29. Examine whether the following results of a piece of computation for obtaining the
second central moment are consistent or not ; N = 120, ∑ f X = –125, ∑ f X2 = 128.
Solution. We have
2
∑ f X2
∑ f X 2 128
μ2 =
–
= 120 – –125
= 1·0670 – 1·0816 = – 0·0146,
120
N
N
which is impossible, since variance cannot be negative. Hence, the given data are inconsistent.
(
)
( )
EXERCISE 7·3.
1. (a) Distinguish between “Skewness” and “Kurtosis” and bring out their importance in describing frequency
distributions.
(b) “Averages, dispension, skewness and kurtosis are complementary to one another in understanding a frequency
distribution.” Elucidate.
SKEWNESS, MOMENTS AND KURTOSIS
7·31
2. (a) Explain the terms Skewness and Kurtosis used in connection with the frequency distribution of a continuous
variable. Give the different measures of skewness (any three of the measures to be given) and kurtosis.
(b) Define skewness and describe briefly the various tests of skewness. [Himachal Pradesh Univ. B.Com., 1998]
(c) Explain briefly how the measures of skewness and kurtosis can be used in describing a frequency distribution.
(d) What do you mean by ‘skewness’ ? How is skewness different from kurtosis ?
3. (a) Explain the relation between the central moments and raw moments of a frequency distribution and hence
express the first four central moments in terms of the raw moments.
(b) What are the raw and central moments of a distribution ? Show that the central moments are invariant under
change of origin but not of scale.
4(a) Define moments. “A frequency distribution can be described almost completely by the first four moments and
two measures based on moments.” Examine the statement.
[Delhi Univ. B.Com. (Hons.), 1996]
(b) Define moments. How are moments helpful in the study of the different aspects of the formation of a
frequency distribution ?
[Delhi Univ. B.Com. (Hons.) (External), 2006]
5. Indicate whether two distributions with the same means, standard deviations and coefficients of skewness must
have same peakedness.
[Delhi Univ. B.Com. (Hons.), 1999]
6. Prove that for any frequency distribution :
(i) Kurtosis is greater than unity.
(ii) Quartile coefficient of skewness is less than 1 numerically.
7. Prove any two of the following :
(i) The sum of squares of deviations is the least when deviations are taken from the mean.
(ii) The standard deviation is affected by the change of scale but not by the change of origin.
(iii) The moment coefficient of kurtosis is not affected by the change of scale.
[Delhi Univ. B.A. (Econ. Hons.), 1991]
8. Prove that the moment coefficient of kurtosis is not affected by the change of scale.
[Delhi Univ. B.A. (Econ. Hons.),1996]
9. Define Pearsonian coefficients β1 and β2, and discuss their utility in Statistics. Define moment coefficient of
skewness. Discuss its utility and limitations, if any.
10. (a) Find the first, second, third and fourth central moments of the set of numbers 2, 4, 6, 8.
Ans. μ 1 = 0, μ2 = 5, μ3 = 0, μ4 = 41.
(b) Given the following data, compute the moment coefficient of skewness and comment on the result.
X:
2
3
7
8
10
[Delhi Univ. B.A. (Econ. Hons.), 1991]
—
–
–
Hint. X = 6; μ 2 = 15 ∑ (x – x )2 = 46
= 9·2 ; μ3 = 15 ∑ (x – x )3 = –18
= –3·6
5
5
β1 = 0·0166 ⇒ γ1 = ⎯√⎯⎯⎯⎯
0·0166 = – 0·12 (Negative sign is taken because μ3 is negative).
Hence, the distribution is very slightly negatively skewed, or it is approximately symmetrical
~ 0).
(·.· β1 =
11. Calculate β1 and β2 (measures of skewness and kurtosis) for the following frequency distribution and hence
comment on the type of the frequency distribution :
x:
2
3
4
5
6
f:
1
3
7
2
1
Ans. β1 = 0·0204 ; β2 = 3·1080. Distribution is approximately normal.
12. Find the first four moments about the mean for the following distribution.
Height (in inches) :
60—62
63—65
66—68
69—71
72—74
Frequency
:
5
18
42
27
8
Ans. μ 1 = 0, μ2 = 8·5275, μ3 = –2·6933, μ4 = 199·3759.
13. Find the variance, skewness and kurtosis of the following distribution by the method of moments :
Class interval
:
0—10
10—20
20–30
30–40
Frequency
:
1
4
3
2
Ans. σ2 = 84, γ1 = 0·0935, β2 = 2·102.
7·32
BUSINESS STATISTICS
14. Explain Sheppard’s corrections for Grouping Errors. What are the conditions to be satisfied for the application
of Sheppard’s corrections ?
[Delhi Univ. B.Com. (Hons.), (External), 2006]
15. For the following distribution, calculate the first four central moments and two beta co-efficients :
Class Interval
:
20—30
30—40
40—50
50—60
60—70
70—80
80—90
Frequency
:
5
14
20
25
17
11
8
[Delhi Univ. B.Com. (Hons.), 2001]
Ans. μ 1 = 0, μ2 = 254, μ3 = 540, μ4 = 1,49,000; β1 = 0·0178, β2 = 2·3095.
16. The first two moments of a distribution about the value 5 of the variable are 2 and 20. Find the mean and the
variance.
Ans. Mean = 7, Variance = 16.
17. The first three moments of a distribution about the value 2 are 1, 22 and 10. Find its mean, standard deviation
and the moment measure of skewness.
Ans. Mean = 3, σ = 4·58, γ1 = – 0·5614.
18. If the first three moments about the origin for a distribution are 10,225 and 0 respectively, calculate the fist
three moments about the value 5 for the distribution.
[Delhi Univ. B.A. (Econ. Hons.), 2005]
Ans. 5, 150, – 2750.
19. For a distribution, the mean is 10, variance is 16, γ 1 is +1 and β2 is 4. Obtain the first four moments about the
origin, i.e., zero. Comment upon the nature of distribution.
[Delhi Univ. B.Com. (Hons.), 2005]
Ans. Moments about origin : μ1′ = 10, μ2′ = 116, μ3′ = 1544, μ4′ = 23184. The distribution is positively skewed
and leptokurtic.
20. The arithmetic mean of a certain distribution is 5. The second and the third moments about the mean are 20
and 140 respectively. Find the third moment of the distribution about 10.
Ans. μ 3′ (about 10) = –285.
21. In a certain distribution the first four moments about the point 4 are 1·5, 17, –30 and 108 respectively. Find the
kurtosis of the frequency curve and comment on its shape.
Ans. β2 = 2·3088. Distt. is Platy-kurtic.
22. The first four moments of a distribution about the value 4 are – 1·5, 17, – 30 and 108 respectively. Calculate :
(i) Moments about mean; (ii) Skewness; (iii) Show whether the distribution is Leptokurtic or Platykurtic.
[Delhi Univ. B.Com. (Hons.), 2009]
Ans. (i) μ1 = 0, μ2 = σ2 = 14·75, μ3 = 39·75, μ 4 = 142·3125.
(ii) Skewness (γ 1) = (μ3/σ3) = 0·7018
(iii) β2 = (μ4/μ22) = 0·654 < 3 ⇒ Distribution is platykurtic.
Note . See Example 7·28.
23. The first four central moments are 0, 4, 8 and 144. Examine the skewness and kurtosis.
Ans. γ1 = 1, β2 = 9
⇒
γ2 = 6.
24. The central moments of a distribution are given by : μ2 = 140, μ3 = 148, μ4 = 6030.
Calculate the moment measures of skewness and kurtosis and comment on the shape of the distribution.
Ans. γ1 = 0·0893 ; β2 = 0·3076 ; Distribution is approximately symmetrical and platy-kurtic.
25. The first four moments of a distribution about value 2 are 1, 2·5, 5·5 and 16 respectively. Calculate the four
moments about mean and comment on the nature of the distribution.
[Delhi Univ. B.Com. (Hons.), 2002]
26. The first four moments of a distribution about the value 3 are 1, 2·5, 5·5 and 16 respectively. Do you think that
the distribution is leptokurtic ?
[Delhi Univ. B.A. (Econ. Hons.), 2009]
Ans. μ2 = 1·5, μ 3 = 0, μ4 = 6 ; β2 = 2·67 < 3. Distribution is platy-kurtic and not leptokurtic.
27. The first four moments of a distribution about the value 5 are equal to 2, 20, 40 and 50. Obtain the mean,
⎯
√
[Delhi Univ. B.A. (Econ. Hons.), 1993]
variance, β1 and β2 for the distribution and comment.
Ans. Mean = 7, Variance = μ2 = 16; μ3 = – 64, μ4 = 162.
μ3
μ4
γ 1 = β1 = 3/2 = – 64 = –1. (Negatively skewed) ; β 2 = 2 = 162
= 0·63 (Platy-kurtic)
μ2
64
μ2 256
⎯√
7·33
SKEWNESS, MOMENTS AND KURTOSIS
28. The following data are given to an economist for the purpose of economic analysis. The data refer to the length
of a certain type of batteries.
N = 100, ∑ fd = 50, ∑ fd2 = 1970, ∑ fd3 = 2948 and ∑ fd4 = 86,752, in which d = (X – 48). Do you think that the
distribution is platykurtic ?
[Delhi Univ. B.Com. (Hons.), 1998]
Ans. β2 = 2·214; β2 < 3, distribution is platy-kurtic.
29. The standard deviation of a symmetrical distribution is 5. What must be the value of the fourth moment about
the mean in order that the distribution be (a) lepto-kurtic, (b) meso-kurtic, and (c) platy-kurtic ?
Ans. Distribution is (a) lepto-kurtic if μ4 > 1875; (b) meso-kurtic if μ4 = 1875; (c) platy-kurtic if μ 4 < 1875.
30. From the data given below, first calculate the first four moments about an arbitrary value and then after
Sheppard’s corrections, calculate the first four moments about the mean. Also calculate β2 and comment on its value.
Average number of hours worked per week by workers in 100 industries in 1998.
Hours worked
:
30—33
33—36
36—39
39–42
No. of Industries
:
2
4
26
47
42—45
15
45—48
6
Ans. Moments after applying Sheppard’s correction are :
μ1 = 0, μ2 = 8·01, μ3 = –20·69, μ4 = 249·393.
31. (a) The first four moments of a distribution about the value 4 are –1·5, 17, –130, 108. Find whether the data are
consistent.
Ans. μ 4 = 108 – 780 + 229·5 – 15·1875 = – 457·6875. Since μ4 is negative, data are inconsistent.
(b) The first four moments of a distribution about 3 are 1, 3, 6 and 8. Is the data consistent ? Explain
[Delhi Univ. B.A. (Econ. Hons.), 2009]
Ans. Inconsistent, because μ4 = – 1 (< 0), which is not possible.
(c) For a distribution, the first four moments about zero are 2, 5, 16 and 30. Is the data consistent ? Explain.
[Delhi Univ. B.A. (Econ. Hons.), 2007]
Ans. Data are inconsistent because μ4 = – 26 (Negative), which is not possible.
32. (a) The standard deviation of a symmetrical distribution is given to be 5. What must be the value of the fourth
moment about the mean in order that the distribution is meso-kurtic ?
[Delhi Univ. B.A. (Econ. Hons.), 1999]
(b) The standard deviation of symmetrical distribution is 3. What must be the value of the 4th moment about the
mean in order that the distribution be mesokurtic ?
[Maharishi Dayanand Univ. M.Com., 1999]
Ans. (a) = 1875, (b) 243.
(c) Comment on the following :
The standard deviation of a symmetrical distribution is 3 and the value of fourth moment about mean is 243 in
order that the distribution is mesokurtic.
[Delhi Univ. B.Com. (Hons.), 2009]
Ans. True; because for Mesokurtic distribution : β2 = 3.
33. Fill in the blanks :
(a) (i) If β2 = 3, the curve is called ……
(b) (i)
(ii) If β2 > 3, the curve is called ……
(ii)
(iii) If β2 < 3, the curve is called ……
(iii)
(iv) If β1 = 0, the curve is called ……
(iv)
Comment on the result when equality sign holds.
Ans. (a) (i) Mesokurtic,
(ii) Leptokurtic,
(iii) Platykurtic
(b) (i) β1 ≥ 0,
(ii) μ4 ≥ 0,
(iii) μ2 ≥ 0,
34. Fill in the blanks :
(i) Literal meaning of skewness is “.......”
(ii) Kurtosis is a measure of …… of the frequency curve.
(iii) For a symmetrical distribution, mean, median and mode ……
(iv) If Mean < Mode, the distribution is …… skewed.
(v) If Mean > Median, the distribution is …… skewed.
(vi) Bowley’s coefficient of skewness lies between …… and ……
(vii) β1 = 0 implies that distribution is ……
(viii) If γ1 > 0, the distribution is …… skewed.
β1 is always … (≥ , = , ≤).
μ4 is always … (≥, = , ≤).
μ2 is always … (≥, = , ≤).
β2 is always … (≥, = , ≤)
(iv) Symmetrical
(iv) β2 ≥ 1.
7·34
BUSINESS STATISTICS
If γ1 < 0, the distribution is …… skewed.
For a normal curve β2 ……
If β2 > 3, the curve is called ……
If β2 < 3, the curve is called ……
An absolute measure of skewness based on moments is ……
An absolute measure of skewness based on quartiles is ……
Relative measure of skewness based on mean, s.d. and mode is ……
Relative measure of kurtosis is ……
Relative measure of skewness in terms of moments is ……
For a moderately asymmetrical distribution :
Mean – Median = ? (Mean – Mode)
If μ3 = –1·48, the curve of the given distribution is stretched more to the ……than to the ……
For a symmetrical distribution, …… are equidistant from ……
Mean = First moment about ……
Variance = …… moment about mean.
If μ1′ is the first moment about the point ‘A’, then Mean = ……
μ2
…
(xxiv) β1 = 3
;
β2 = 2
γ 1 = ……
;
γ 2 = ……
……
μ2
(xxv) In a moderately asymmetrical distribution the distance between … and … is … the distance between … and
…
Ans.
(i) Lack of symmetry,
(ii) Convexity (flatness or
(iii) Coincide,
(iv) Negatively,
peakedness),
(v) Positively,
(vi) –1 and 1,
(vii) Symmetrical,
(viii) Positively,
(ix) Negatively,
(x) 3,
(xi) Lepto-kurtic,
(xii) Platy-kurtic,
(ix)
(x)
(xi)
(xii)
(xiii)
(xiv)
(xv)
(xvi)
(xvii)
(xviii)
(xix)
(xx)
(xxi)
(xxii)
(xxiii)
(xii) μ3,
(xvii) β1 or γ 1,
(xxi) Origin (i.e., A = 0),
(xiv) (Q3 – Md) – (Md – Q1),
(xviii) 1/3,
(xxii) Second,
(xv) (M – Mo)/σ,
(xix) Left, Right,
(xxiii) A + μ1′,
(xvi) β2 or γ 2,
(xx) Quartiles, median,
μ3
(xxiv) μ23, μ4, β1 = 3/2 ,
μ2
γ 2 = β2 – 3,
⎯√
(xxv) Mean, Mode, Thrice, Mean, Median.
35. State whether the following statements are true (T) or false (F).
(i) Skewness studies the flatness or peakedness of the distribution.
(ii) Kurtosis means ‘lack of symmetry’.
(iii) For a symmetrical distribution β1 = 0.
(iv) Skewness and kurtosis help us in studying the shape of the frequency curve.
(v) Bowley’s coefficient of skewness lies between –3 and 3.
(vi) Two distributions having the same values of mean, s.d. and skewness must have the same kurtosis.
(vii) A positively skewed distribution curve is stretched more to the right than to the left.
(viii) If β2 > 3, the curve is called platy-kurtic.
(ix) If β2 = 3, the curve is called normal.
(x) β1 is always non-negative.
β2 can be negative.
Variance = μ2 (2nd moment about mean).
For a symmetrical distribution : μ1 = μ3 = μ5 = …… = 0
For a symmetrical distribution : Mean > Median > Mode.
μ4
(xv) β1 = 2
μ3
Ans.
(i) F,
(ii) F,
(iii) T,
(vi) F,
(vii) T,
(viii) F,
(xi) F,
(xii) T,
(xiii) T,
(xi)
(xii)
(xiii)
(xiv)
(iv) T,
(ix) T,
(xiv) F,
(v) F
(x) T,
(xv) F.
8
Correlation Analysis
8·1. INTRODUCTION
So far we have confined our discussion to univariate distributions only i.e., the distributions involving
only one variable and also saw how the various measures of central tendency, dispersion, skewness and
kurtosis can be used for the purposes of comparison and analysis. We may, however, come across certain
series where each item of the series may assume the values of two or more variables. If we measure the
heights and weights of n individuals, we obtain a series in which each unit (individual) of the series
assumes two values—one relating to heights and the other relating to weights. Such distribution, in which
each unit of the series assumes two values is called a bivariate distribution. Further, if we measure more
than two variables on each unit of a distribution, it is called a multivariate distribution. In a series, the units
on which different measurements are taken may be of almost any nature such as different individuals,
times, places, etc. For example we may have :
(i) The series of marks of individuals in two subjects in an examination.
(ii) The series of sales revenue and advertising expenditure of different companies in a particular year.
(iii) The series of exports of raw cotton in crores of rupees and imports of manufactured goods during
number of years from 1989 to 1994, say.
(iv) The series of ages of husbands and wives in a sample of selected married couples and so on.
Thus in a bivariate distribution we are given a set of pairs of observations, one value of each pair being
the values of each of the two variables.
In a bivariate distribution, we may be interested to find if there is any relationship between the two
variables under study. The correlation is a statistical tool which studies the relationship between two
variables and correlation analysis involves various methods and techniques used for studying and
measuring the extent of the relationship between the two variables.
WHAT THEY SAY ABOUT CORRELATION — SOME DEFINITIONS AND USES
“When the relationship is of a quantitative nature, the appropriate statistical tool for
discovering and measuring the relationship and expressing it in a brief formula is known as
correlation.”—Croxton and Cowden
“Correlation is an analysis of the covariation between two or more variables.” —A.M. Tuttle
“Correlation analysis contributes to the understanding of economic behaviour, aids in
locating the critically important variables on which others depend, may reveal to the economist
the connections by which disturbances spread and suggest to him the paths through which
stabilising forces may become effective.”—W.A. Neiswanger
“The effect of correlation is to reduce the range of uncertainty of our prediction.” —Tippett
Two variables are said to be correlated if the change in one variable results in a corresponding
change in the other variable.
8·1·1. Types of Correlation
(a) POSITIVE AND NEGATIVE CORRELATION
If the values of the two variables deviate in the same direction i.e., if the increase in the values of one
variable results, on an average, in a corresponding increase in the values of the other variable or if a
8·2
BUSINESS STATISTICS
decrease in the values of one variable results, on an average, in a corresponding decrease in the values of
the other variable, correlation is said to be positive or direct.
Some examples of series of positive correlation are :
(i) Heights and weights.
(ii) The family income and expenditure on luxury items.
(iii) Amount of rainfall and yield of crop (up to a point).
(iv) Price and supply of a commodity and so on.
On the other hand, correlation is said to be negative or inverse if the variables deviate in the opposite
direction i.e., if the increase (decrease) in the values of one variable results, on the average, in a
corresponding decrease (increase) in the values of the other variable.
Some examples of negative correlation are the series relating to :
(i) Price and demand of a commodity.
(ii) Volume and pressure of a perfect gas.
(iii) Sale of woollen garments and the day temperature, and so on.
(b) LINEAR AND NON-LINEAR CORRELATION
The correlation between two variables is said to be linear if corresponding to a unit change in one
variable, there is a constant change in the other variable over the entire range of the values. For example, let
us consider the following data :
x
1
2
3
4
5
y
5
7
9
11
13
Thus for a unit change in the value of x, there is a constant change viz., 2 in the corresponding values of
y. Mathematically, above data can be expressed by the relation
y = 2x + 3
In general, two variables x and y are said to be linearly related, if there exists a relationship of the form
y = a + bx
… (*)
between them. But we know that (*) is the equation of a straight line with slope ‘b’ and which makes an
intercept ‘a’ on the y-axis [c.f. y = m x + c form of equation of the line]. Hence, if the values of the two
variables are plotted as points in the xy-plane, we shall get a straight line. This can be easily checked for the
example given above. Such phenomena occur frequently in physical sciences but in economics and social
sciences, we very rarely come across the data which give a straight line graph. The relationship between
two variables is said to be non-linear or curvilinear if corresponding to a unit change in one variable, the
other variable does not change at a constant rate but at fluctuating rate. In such cases if the data are plotted
on the xy-plane, we do not get a straight line curve. Mathematically speaking, the correlation is said to be
non-linear if the slope of the plotted curve is not constant. Such phenomena are common in the data relating
to economics and social sciences.
Since the techniques for the analysis and measurement of non-linear relation are quite complicated and
tedious as compared to the methods of studying and measuring linear relationship, we generally assume
that the relationship between the two variables under study is linear. In this chapter, we shall confine
ourselves to the measurement of linear relationship only. The measurement of non-linear relationship is,
however, beyond the scope of this book.
Remark. The study of correlation is easy in physical sciences since on the basis of experimental
results, it is easy to establish mathematical relationship between two or more variables under study. But in
social and economic sciences, it is very difficult to establish mathematical relationship between the
variables under study since in such phenomena, the values of the variables under study are affected
simultaneously by multiplicity of factors and it is extremely difficult, sometimes impossible, to study the
effects of each factor separately. Hence, in the data relating to social and economic phenomena, the study
of correlation cannot be as accurate and precise.
8·1·2. Correlation and Causation. Correlation analysis enables us to have an idea about the degree
and direction of the relationship between the two variables under study. However, it fails to reflect upon the
CORRELATION ANALYSIS
8·3
cause and effect relationship between the variables. In a bivariate distribution, if the variables have the
cause and effect relationship, they are bound to vary in sympathy with each other and, therefore, there is
bound to be a high degree of correlation between them. In other words, causation always implies
correlation. However, the converse is not true i.e., even a fairly high degree of correlation between the two
variables need not imply a cause and effect relationship between them. The high degree of correlation
between the variables may be due to the following reasons :
1. Mutual dependence. The phenomena under study may inter-influence each other. Such situations are
usually observed in data relating to economic and business situations. For instance, it is well-known
principle in economics that prices of a commodity are influenced by the forces of supply and demand. For
instance, if the price of a commodity increases, its demand generally decreases (other factors remaining
constant). Here increased price is the cause and reduction in demand is the effect. However, a decrease in
the demand of a commodity due to emigration of the people or due to fashion or some other factors like
changes in the tastes and habits of people may result in decrease in its price. Here, the cause is the reduced
demand and the effect is the reduced price. Accordingly, the two variables may show a good degree of
correlation due to interaction of each on the other, yet it becomes very difficult to isolate the exact cause
from the effect.
2. Both the variables being influenced by the same external factors. A high degree of correlation
between the two variables may be due to the effect or interaction of a third variable or a number of
variables on each of these two variables. For example, a fairly high degree of correlation may be observed
between the yield per hectare of two crops, say, rice and potato, due to the effect of a number of factors like
favourable weather conditions, fertilizers used, irrigation facilities, etc., on each of them. But none of the
two is the cause of the other.
3. Pure chance. It may happen that a small randomly selected sample from a bivariate distribution may
show a fairly high degree of correlation though, actually, the variables may not be correlated in the
population. Such correlation may be attributed to chance fluctuations. Moreover, the conscious or
unconscious bias on the part of the investigator, in the selection of the sample may also result in high
degree of correlation in the sample. In this connection, it may be worthwhile to make a mention of the two
phenomena where a fairly high degree of correlation may be observed, though it is not possible to conceive
them as being causally related. For example, we may observe a high degree of correlation between the size
of shoe and the intelligence of a group of persons. Such correlation is called spurious or non-sense
correlation. [For details see § 8·4·2 (iii).]
8·2. METHODS OF STUDYING CORRELATION
We shall confine our discussion to the methods of ascertaining only linear relationship between two
variables (series). The commonly used methods for studying the correlation between two variables are :
(i) Scatter diagram method.
(ii) Karl Pearson’s coefficient of correlation (Covariance method).
(iii) Two-way frequency table (Bivariate correlation method).
(iv) Rank method.
(v) Concurrent deviations method.
8·3. SCATTER DIAGRAM METHOD
Scatter diagram is one of the simplest ways of diagrammatic representation of a bivariate distribution
and provides us one of simplest tools of ascertaining the correlation between two variables. Suppose we are
given n pairs of values (x1, y 1 ), (x2, y 2 ), …, (xn, y n ) of two variables X and Y. For example, if the variables X
and Y denote the height and weight respectively, then the pairs (x1 , y 1 ), (x 2 , y 2 ), … , (xn, y n ) may represent
the heights and weights (in pairs) of n individuals. These n points may be plotted as dots (.) on the x-axis
and y-axis in the xy-plane. (It is customary to take the dependent variable along the y-axis and independent
variable along the x-axis.) The diagram of dots so obtained is known as scatter diagram. From scatter
diagram we can form a fairly good, though rough, idea about the relationship between the two variables.
The following points may be borne in mind in interpreting the scatter diagram regarding the correlation
between the two variables :
8·4
BUSINESS STATISTICS
(i) If the points are very dense i.e., very close to each other, a fairly good amount of correlation may be
expected between the two variables. On the other hand, if the points are widely scattered, a poor correlation
may be expected between them.
(ii) If the points on the scatter diagram reveal any trend (either upward or downward), the variables are
said to be correlated and if no trend is revealed, the variables are uncorrelated.
(iii) If there is an upward trend rising from lower left hand corner and going upward to the upper right
hand corner, the correlation is positive since this reveals that the values of the two variables move in the
same direction. If, on the other hand, the points depict a downward trend from the upper left hand corner to
the lower right hand corner, the correlation is negative since in this case the values of the two variables
move in the opposite directions.
(iv) In particular, if all the points lie on a straight line starting from the left bottom and going up
towards the right top, the correlation is perfect and positive, and if all the points lie on a straight line
starting from left top and coming down to right bottom, the correlation is perfect and negative.
The following diagrams of the scattered data depict different forms of correlation.
PERFECT POSITIVE
CORRELATION
PERFECT NEGATIVE
CORRELATION
LOW DEGREE OF
POSITIVE CORRELATION
Y
Y
•
•
•
•
•
•
Y
•
•
•
•
•
O
O
X
•
•
•
•
•
•
•
•
•
•
•
Fig. 8.3.
HIGH DEGREE OF
POSITIVE CORRELATION
Y
•
X
Fig. 8.2.
LOW DEGREE OF
NEGATIVE CORRELATION
••
• •• •
•••••
•
••
• •• •
••
••• •
•
HIGH DEGREE OF
NEGATIVE CORRELATION
Y
••
•• •
• ••
• •
•
X O
Fig. 8.4.
NO CORRELATION
Y
X
Fig. 8.5.
O
X
Fig. 8.7.
••
•• •
• ••
O
X
Fig. 8.6.
NO CORRELATION
Y
• • • ••
••
•
•
•
•
•
•
•
•
•
•
•
•
•• • • •
• •
•
O
X
Fig. 8.1.
O
•
• •
• • •
•
•
•
•
•
•
Y
• •
•
•
•
•
•
•
•
•
•
•
•
•
••
•• • •
•
•
•
•
•
•
•
•
•
O
X
Fig. 8.8.
8·5
CORRELATION ANALYSIS
Remarks 1. The method of scatter diagram is readily comprehensible and enables us to form a rough
idea of the nature of the relationship between the two variables merely by inspection of the graph.
Moreover, this method is not affected by extreme observations whereas all mathematical formulae of
ascertaining correlation between two variables are affected by extreme observations. However, this method
is not suitable if the number of observations is fairly large.
2. The method of scatter diagram only tells us about the nature of the relationship whether it is positive
or negative and whether it is high or low. It does not provide us an exact measure of the extent of the
relationship between the two variables.
3. The scatter diagram enables us to obtain an approximate estimating line or line of best fit by free
hand method. The method generally consists in stretching a piece of thread through the plotted points to
locate the best possible line. However, a rigorous method of obtaining the line of best fit is discussed in
next chapter (Regression Analysis).
Example 8·1. Following are the heights and weights of 10 students of a B.Com. class.
Height (in inches ) X : 62
72
68
58
65
70
66
63
60
72
Weight (in kgs.)
Y : 50
65
63
50
54
60
61
55
54
65
Draw a scatter diagram and indicate whether the correlation is positive or negative.
Solution. The scatter diagram of the above data is shown below.
SCATTER DIAGRAM
Y
66
62
60
58
56
54
52
•
•
•
•
•
•
•
•
•
—
–
—
–
—
–
—
–
—
–
—
–
—
–
—
50
—
–
—
–
—
–
—
–
—
–
—
–
—
–
—
–
—
64
0 58
60
62
64
66
68
70
72
X
Fig. 8.9.
Since the points are dense i.e., close to each other, we may expect a high degree of correlation between
the series of heights and weights. Further, since the points reveal an upward trend starting from left bottom
and going up towards the right top, the correlation is positive. Hence, we may expect a fairly high degree of
positive correlation between the series of heights and weights in the class of B.Com. students.
EXERCISE 8·1
1. Explain the concept of correlation. Clearly explain with suitable illustrations its role in dealing with business
problems.
2. (a) Define correlation. Explain various types of correlation with suitable examples.
[Delhi Univ. B.Com. (Pass), 2000]
(b) State the nature of the following correlations (positive, negative or no correlation) :
(i) Sale of woollen garments and the day temperature;
(ii) The colour of the saree and the intelligence of the lady who wears it ; and
(iii) Amount of rainfall and yield of crop.
8·6
BUSINESS STATISTICS
3. Define correlation. Discuss its significance. Does correlation always signify causal relationship between two
variables ? Explain with illustration.
4. (a) Does the high degree of correlation between the two variables signify the existence of cause and effect
relationship between the two variables ?
(b) Does correlation imply causation between two variables ?
[Delhi Univ. B.Com. (Hons.), 2008]
5. (a) What is ‘spurious correlation’ and ‘non-sense or chance correlation’ ? Explain with the help of an
example.
[Delhi Univ. B.Com. (Pass), 1997]
(b) Comment on the following statement : “A high degree of positive correlation between the ‘size of the shoe’
and the ‘intelligence’ of a group of individuals implies that people with bigger shoe size are more intelligent than the
people with lower shoe size”.
6. How far do you agree with the conclusion drawn in the following case ? Why ?
Two series — quantity of money in circulation and general price index — are found to possess positive correlation
of a fairly high order. From this, it is concluded that one is the cause and the other the effect in a direct causal
relationship.
7. (a) Distinguish clearly between :
(i) Positive and Negative correlation ;
(ii) Linear and non-linear correlation.
(b) “If the two or more quantities vary in sympathy so that movements in one tend to be accompanied by
corresponding movements in other(s), then they are said to be correlated.” Discuss.
8. What is correlation ? What is a scatter diagram ? How does it help in studying correlation between two
variables, in respect of both its nature and extent ?
9. (a) What is correlation ? Explain the implications of positive and negative correlation. Show by means of scatter
diagram, the presence of perfect positive and perfect negative correlation.
[C.A. (Foundation), May 1996]
(b) Illustrate a perfect negative correlation on a scatter diagram.
[Delhi Univ. B.Com. (Pass), 1998]
10. (a) What is a scatter diagram ? How is it useful in the study of correlation between two variables ? Explain
with suitable examples.
[Delhi Univ. B.Com. (Hons.), (External), 2006]
(b) Write a note on scatter diagram. Draw sketches of scatter diagram to show the following correlation between
two variables x and y :
(i) linear ;
(ii) linear and perfect;
(iii) non-linear;
(iv) x and y uncorrelated.
(c) While drawing a scatter diagram, if all points appear to form a straight line going downward from left to right,
then it is inferred that there is :
(i) Perfect positive correlation ;
(ii) Simple positive correlation;
(iii) Perfect negative correlation ;
(iv) No correlation.
Ans. (iii)
11. Given the following pairs of values :
Capital employed (Crores of Rs.) :
2
3
5
6
8
9
Profits (Lakhs of Rs.)
:
6
5
7
8
12
11
(a) Make a scatter diagram.
(b) Do you think that there is any correlation between profits and capital employed ? Is it positive or negative ? Is
it high or low ?
(c) By graphic inspection, draw an estimating line.
12. “Even a high degree of correlation does not mean that a relationship of cause and effect exists between the two
correlated variables”. Discuss.
13. Draw a scatter diagram from the following data :
Height (inches)
:
62
72
70
60
67
70
64
65
60
70
Weight (lbs.)
:
50
65
63
52
56
60
59
58
54
65
Also indicate whether correlation is positive or negative.
Ans. Positive Correlation.
14. Draw a scatter diagram for the data given below and interpret it.
X:
10
20
30
40
50
60
70
80
Y:
32
20
24
36
40
28
38
44
8·7
CORRELATION ANALYSIS
8·4. KARL PEARSON’S COEFFICIENT OF CORRELATION (COVARIANCE METHOD)
A mathematical method for measuring the intensity or the magnitude of linear relationship between
two variable series was suggested by Karl Pearson (1867-1936), a great British Bio-metrician and
Statistician and is by far the most widely used method in practice.
Karl Pearson’s measure, known as Pearsonian correlation coefficient between two variables (series) X
and Y, usually denoted by r (X, Y) or rxy or simply r, is a numerical measure of linear relationship between
them and is defined as the ratio of the covariance between X and Y, written as Cov (x, y), to the product of
the standard deviations of X and Y. Symbolically,
Cov (x, y)
r =
… (8·1)
σx σy
If (x 1 , y1), (x 2 , y2), … , (xn, yn) are n pairs of observations of the variables X and Y in a bivariate
distribution, then
Cov (x, y) = 1 ∑(x – x– ) (y – –y) ; σx =
n
1
∑(x – x– )2 ; σ y =
n
1
∑(y – –y )2
n
summation being taken over n pairs of observations. Substituting in (8·1), we get
1
∑ (x – –x ) (y – –y )
∑ (x – –x ) (y – –y )
n
r =
=
1
1
∑ (x – –x )2 ∑ (y – –y )2
∑
(x – –x )2. ∑(y – –y )2
n
n
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
The formula (8·3) can also be written as :
r=
∑ dx dy
∑ dx2 . ∑ dy2
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
… (8·2)
… (8·3)
… (8·3a)
where dx and dy denote the deviations of x and y values from their arithmetic means –x and –y respectively
i.e.,
dx = x – –x , dy = y – –y
… (8·3b)
Simplifying (8·2), we get
1
Cov (x, y) = ∑ xy – –x –y
… (8·4)
n
∑x ∑y
1
1
⇒
Cov (x, y) = ∑ xy –
= 2 n∑ xy – ∑ x ∑ y
… (8·4a)
n
n
n
n
( )( )
( )( )]
[
1
∑(x – –x ) 2 = n1 ∑ x2 – –x 2
n
∑x 2 1
1
= ∑x2 –
= 2 n∑ x2 –
n
n
n
σx2 =
( )
[c.f. Chapter 6]
( ∑x ) ]
[ n∑ y – ( ∑ y ) ]
[
2
… (8·4b)
2
1
2
2
n
Substituting from (8·4a), (8·4b) and (8·4c) in (8·1), we get
1
n∑ xy – ∑ x ∑ y
n2
r =
2
1
1
n∑ x2 – ∑ x
.
n∑ y2 –
n2
n2
σy2 =
Similarly, we have,
( ) ( )]
[
( )}
(∑ y ) }
{
n ∑ xy – (∑ x ) ( ∑ y )
[ n ∑ x – ( ∑ x ) ] [ n∑ y – ( ∑ y ) ]
{
=
… (8·4c)
2
2
2
2
2
… (8·5)
8·8
BUSINESS STATISTICS
Remarks. Formula (8·3) or (8·3a) is quite convenient to apply if the means –x and y– come out to be
integers (i.e., whole numbers). If –x or /and –y is (are) fractional then the formula (8·3) or (8·3a) is quite
cumbersome to apply, since the computations of ∑(x – x– )2, ∑ (y – –y )2 and ∑(x – x– ) (y – –y ) are quite time
consuming and tedious. In such a case formula (8·5) may be used provided the values of x or/and y are
small. But if x and y assume large values, the calculation of ∑ x 2 , ∑ y 2 and ∑ xy is again quite time
consuming.
Thus if (i) –x and –y are fractional and (ii) x and y assume large values, the formulae (8·3) and (8·5) are
not generally used for numerical problems. In such cases, the step deviation method where we take the
deviations of the variables X and Y from any arbitrary points, is used. We shall discuss this method after the
properties of correlation coefficient (c.f. § 8·4·1).
2. Karl Pearson’s correlation coefficient is also known as the product moment correlation coefficient.
Example 8·2. Calculate Karl Pearson’s coefficient of correlation between expenditure on advertising
and sales from the data given below.
Advertising expenses (’000 Rs. ) : 39
65
62
90
82
75
25
98
36
78
Sales (lakh Rs.)
: 47
53
58
86
62
68
60
91
51
84
Solution. Let the advertising expenses (in ’000 Rs.) be denoted by the variable x and the sales (in lakh
Rs.) be denoted by the variable y.
x
y
39
65
62
90
82
75
25
98
36
78
47
53
58
86
62
68
60
91
51
84
∑x = 650
∑y = 660
CALCULATIONS FOR CORRELATION COEFFICIENT
–
–
dy2
dx = x – x = x – 65 dy = y – y = y – 66
dx2
– 26
0
– 3
25
17
10
– 40
33
– 29
13
∑ dx = 0
– 19
– 13
– 8
20
– 4
2
– 6
25
– 15
18
∑ dy = 0
dxdy
676
0
9
625
289
100
1600
1089
841
169
361
169
64
400
16
4
36
625
225
324
494
0
24
500
– 68
20
240
825
435
234
∑ dx2 = 5398
∑ dy2 = 2224
∑ dxdy = 2704
–x = ∑x = 650 = 65 ; –y = ∑y = 660 = 66 ;
∴ dx = x – –x = x – 65 ; dy = y – –y = y – 66
10
10
n
n
∑dxdy
2704
2704
2704
r =
=
=
= 3464·8451 = 0·7804
2
2
⎯
√
⎯⎯⎯⎯⎯⎯⎯
5398
×
2224
⎯
√
⎯⎯⎯⎯⎯⎯⎯
12005152
⎯⎯⎯⎯⎯⎯⎯⎯
√
∑dx . ∑dy
Aliter.
⇒
log r = log 2704 – 21 [log 5398 + log 2224]
–
= 3·4320 – 12 (3·7325 + 3·3472) = 3·4320 – 3·53985 = – 0·10785 = 1·89215
–
⇒
r = Antilog ( 1·89215) = 0·7802
Hence, there is a fairly high degree of positive correlation between expenditure on advertising and
sales. We may, therefore, conclude that in general, sales have increased with an increase in the advertising
expenses.
Example 8·3. From the following table calculate the coefficient of correlation by Karl Pearson’s
method.
X
6
2
10
4
8
Y
9
11
?
8
7
Arithmetic means of X and Y series are 6 and 8 respectively.
8·9
CORRELATION ANALYSIS
Solution. First of all we shall find the missing value of Y. Let the missing value in Y-series be a. Then
the mean –y is given by :
–y = ∑Y = 9 + 11 + a + 8 + 7 = 35 + a = 8 (given )
5
5
n
⇒
35 + a = 5 × 8 = 40
⇒
a = 40 – 35 = 5
CALCULATION OF CORRELATION COEFFICIENT
—
—
9
11
5
8
7
0
–4
4
–2
2
1
3
–3
0
–1
∑Y = 40
0
0
X
Y
6
2
10
4
8
∑X = 30
We have
X – X = X – 6 (Y – Y ) = Y – 8
30
X = ∑X
5 = 5 = 6,
—
—
(X – X )2
—
(Y – Y )2
0
16
16
4
4
1
9
9
0
1
—
—
—
—
(X – X ) (Y – Y )
0
– 12
– 12
0
–2
—
—
∑(X – X )2 = 40 ∑(Y – Y )2 = 20 ∑(X – X ) (Y – Y ) = – 26
40
Y = ∑Y
5 = 5 =8
—
Karl Pearson’s correlation coefficient is given by :
—
—
∑(X – X ) (Y – Y )
Cov(X‚ Y)
– 26
r=
=
= – 26 = – 26 = 28·2843
= – 0·9192 ~
– – 0·92
σx σ y
— 2
— 2
⎯⎯⎯⎯⎯⎯
√
40 × 20 √
⎯⎯⎯
800
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
∑(X – X ) ∑(Y – Y )
Example 8·4. Calculate the coefficient of correlation between X and Y series from the following data :
Series
X
Y
No. of pairs of observations
15
15
Arithmetic mean
25
18
Standard deviation
3·01
3·03
Sum of squares of deviations from mean
136
138
Summation of product deviations of X and Y series from their respective arithmetic means = 122.
Solution. In the usual notations, we are given :
n = 15, –x = 25, –y = 18,
σ = 3·01, σ = 3·03
x
y
∑(x – –x )2 = 136, ∑(y – –y )2 = 138,
and
∑ (x – –x ) (y – –y ) = 122.
Karl Pearson’s correlation coefficient between X-series and Y-series is given by
∑ (x – –x ) (y ––y )
122
122
r =
= 15 × 3·01
× 3·03 = 136·8049 = 0·8918
n σx σ y
Remark. Here some of the given data are superfluous viz., –x , –y , ∑(x – –x ) 2 , ∑(y – –y ) 2 .
Aliter. We may also compute the correlation coefficient using the formula
∑(x – –x ) (y ––y )
122
r =
= 122
= 122 = 136·9964
= 0·8905
–
–
⎯
√
⎯⎯⎯⎯⎯⎯⎯
136
×
138
⎯
√
⎯⎯⎯
18768
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
∑(x – x )2 ∑(y – y ) 2
If we use this formula, then the data relating to n, –x, –y , σ and σ are superfluous.
x
y
Example 8·5. The coefficient of correlation between two variables X and Y is 0·48. The covariance is
36. The variance of X is 16. Find the standard deviation of Y.
Solution. We are given :
rxy = 0·48, Cov (X, Y) = 36, σx2 = 16
⇒
σx = 4
Cov (X‚ Y)
Cov (X‚Y)
9
We have :
rxy =
⇒
σy =
= 4 ×360·48 = 0·48
= 18·75.
σx σ y
σx rxy
8·10
BUSINESS STATISTICS
Example 8·6. Given the following information :
rxy = 0·8, ∑xy = 60, σ y = 2·5 and
∑x 2 = 90, where x and y are the deviations from the
respective means, find the number of items (n).
[Delhi Univ. B.Com. (Hons.), 2002]
— 2 1
—
1
90
Solution. We have :
σx2 = ∑(X – X ) = ∑x2 =
(·. · X – X = x)
n
n
n
—
—
∑(X – X ) (Y – Y ) ∑ xy
r =
=
n σX σY
nσx σy
⇒
60
0·8 =
n
90
n
=
. (2·5)
60 × 2
⎯n √
√
⎯⎯90 × 5
—
—
[·.· X – X = x ; Y – Y = y ; σX = σx, σY = σy]
=
24
⎯n √
√
⎯⎯90
× 24
n = 90 ×240·8
× 0·8 = 10.
⇒
Example 8.7. The covariance of two perfectly correlated variables X and Y is 96. Determine σX and
σY if it is known that variance of X and that of Y is in the ratio of 4 : 9
.
[Delhi Univ. B.A. (Econ. Hons.), 2008]
σX2 4
σX 2
2
Solution. We are given : Cov (X, Y) = 96 …(1) ;
=
⇒
=
⇒
σ X = σY
…(2)
σY 2 9
σY 3
3
The correlation coefficient r = r (X, Y) is given by
r=
Cov (X‚ Y)
96
144
=2
= 2
σXσY
σY
3 σY . σY
[From (1) and (2)]
…(3)
For perfectly correlated variables, we have r = r (X, Y) = ± 1. But since Cov (X, Y) = 96 > 0, r is also
positive. Hence r = + 1.
144
Substituting in (3) , we get ;
1 = 2
⇒
σY2 = 144
⇒
σY = 12
σY
2
2
Substituting in (2), we get :
σX = σY = × 12 = 8
3
3
∴
σX = 8, σY = 12.
Example 8.8. Given below is the information relating to marks in Statistics (X) and marks in
Accountancy (Y) obtained by the students of a class.
Covariance between X and Y = 144 ; Second moment of X about 20 = 244;
First moment of X about 20 = 10 ; Arithmetic mean of Y = 45
Correlation coefficient between X and Y = 0·75.
Calculate coefficient of variation of marks of Statistics and that of marks of Accountancy. In which
subject the performance of students is more consistent ?
[Delhi Univ. B.Com. (Hons.), (External), 2005]
Solution. We are given : Cov (X, Y) = 144;
μ1 ′ = First moment of X about ‘20’ = 10
μ2 ′ = Second moment of X about ‘20’ = 244
rXY = 0.75 ;
⇒
⇒
—
—
Y = 45
X = A + μ1 ′ = 20 + 10 = 30
μ2 = μ2 ′ – μ1 ′2 = 244 – 100 = 144
…(i)
…(ii)
…(iii)
σX2 = μ2 = 144
⇒
σX = √
⎯⎯⎯
144 = 12
…(iv)
Cov (X, Y)
144
12
rXY =
⇒
0·75 =
⇒
σY =
= 16
…(v)
σX σ Y
12 σY
(3/4)
The coefficient of variation (C.V.) of marks of Statistics and Accountancy are given by :
σ
12
C.V. (Statistics)
= C.V. (X) = —X × 100 = × 100 = 40
[From (iii) and (iv)]
30
X
σ
16
C.V. (Accountancy) = C.V. (Y) = —Y × 100 = × 100 = 35·55
[From (i) and (v)]
45
Y
Since C.V. (Y) < C.V. (X), the performance of the students is more consistent in accountancy.
∴
8·11
CORRELATION ANALYSIS
Example 8·9. A computer while calculating correlation coefficient between two variables X and Y
from 25 pairs of observations obtained the following results :
n = 25, ∑X = 125, ∑X2 = 650, ∑Y = 100, ∑Y2 = 460, ∑XY = 508
It was, however, discovered at the time of checking that two pairs of observations were not correctly
copied. They were taken as (6, 14) and (8, 6) while the correct values were (8, 12) and (6, 8). Prove that the
correct value of the correlation coefficient should be 2 / 3.
[Delhi Univ. B.Com. (Hons.) 2008]
Solution.
Corrected ∑X = 125 – 6 – 8 + 8 + 6 = 125
; Corrected ∑Y = 100 – 14 – 6 + 12 + 8 = 100
Corrected ∑X2 = 650 – 62 – 82 + 8 2 + 62 = 650 ; Corrected ∑Y2 = 460 – 14 2 – 62 + 122 + 82 = 436
Corrected ∑XY = 508 – 6 × 14 – 8 × 6 + 8 × 12 + 6 × 8 = 520
Corrected value of r is given by :
n∑XY –(∑X ) (∑Y)
25 × 520 – 125 × 100
r =
=
2
2
2
2
[n∑X – (∑X) ] × [n∑Y – (∑Y) ] √
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
[25 × 650 – (125)2] × [25 × 436 – (100)2]
=
13000 ± 12500
(16255 ± 15625) × (10900 ± 10000)
=
500
625 × 900
=
500
2
25 × 30 = 3
8·4·1. Properties of Correlation Coefficient
Property I. Limits for Correlation Coefficient
Pearsonian correlation coefficient can not exceed 1 numerically. In other words it lies between –1 and
+1. Symbolically,
–1≤r≤1
… (8·6)
Remarks 1. This result provides us a check on our calculations. If in any problem, the obtained value
of r lies outside the limits ± 1, this implies that there is some mistake in our calculations.
2. r = + 1 implies perfect positive correlation between the variables and r = – 1 implies perfect negative
correlation between the variables.
Property II. Correlation coefficient is independent of the change of origin and scale. Mathematically,
if X and Y are the given variables and they are transformed to the new variables U and V by the change of
origin and scale viz.,
x–A
y–B
u =
and
v =
; h > 0, k > 0
… (8·7)
h
k
where A, B, h and k are constants, h > 0, k > 0; then the correlation coefficient between x and y is same
as the correlation coefficient between u and v i.e.,
r(x, y) = r (u, v)
⇒
rxy = ruv
… (8·7a)
Remark. This is one of the very important properties of the correlation coefficient and is extremely
helpful in numerical computation of r. We had already stated [c.f. Remark to § 8·4] that formulae (8·3) and
(8·5) become quite tedious to use in numerical problems if –x and /or –y are in fractions or if x and y are
large. In such cases we can conveniently change the origin and scale (if possible) in X or/and Y to get new
variables U and V and compute the correlation between U and V by the formula
∑(u – –u ) (v – –v )
n∑uv – (∑u) (∑v)
ruv =
=
… (8·7b)
–
–
2 – (∑u)2 ] [n∑v2 – (∑v )2 ]
2
2
[
n
∑u
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
∑(u – u ) . ∑(v – v )
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
Now using the property II, we finally get :
rxy = ruv .
Property III. Two independent variables are uncorrelated but the converse is not true.
Proof. We have proved in (8·4) that
Cov (x, y) = 1 ∑xy – –x . –y = E(xy) – E(x)E(y),
… (*)
n
because expected value of a variable is nothing but its arithmetic mean. [See Chapter 13 on Expectation].
8·12
BUSINESS STATISTICS
If x and y are independent variables then [c.f. Chapter 13 on Expectation],
E(x . y) = E(x)E(y)
Substituting in (*), we get
Cov (x, y) = E(x)E(y) – E(x)E(y) = 0
Hence, if x and y are independent then
Cov (x‚ y)
0
rxy =
=
=0
σxσy
σxσy
⇒
Independent variables are uncorrelated.
Converse. However, the converse of the theorem is not true i.e., uncorrelated variables need not
necessarily be independent. As an illustration consider the following bivariate distribution.
∴
x
–4
–3
–2
–1
1
2
3
4
∑x = 0
y
16
9
4
1
1
4
9
16
∑y = 60
xy
– 64
– 27
–8
–1
1
8
27
64
∑ xy = 0
rxy =
n∑xy – (∑x )(∑y)
(∑x) √
(∑y )
⎯⎯⎯⎯⎯⎯⎯⎯⎯
⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
n∑x 2 –
2
n∑y 2 –
2
=
8 × 0 – 0 × 60
= 0,
– (∑x )2 √
n∑y 2 – (∑y)2
⎯⎯⎯⎯⎯⎯⎯⎯
√
⎯⎯⎯⎯⎯⎯⎯⎯⎯
n∑x 2
because zero divided by any finite quantity is zero. Hence, in the above example the variables x and y are
uncorrelated. But if we examine the data carefully, we find that x and y are not independent but are
connected by the relation y = x 2 . The above example illustrates that uncorrelated variables need not be
independent.
Remarks 1. One should not be confused with the words of uncorrelation and independence. rxy = 0 i.e.,
uncorrelation between the variables x and y simply implies the absence of any linear (straight line)
relationship between them. They may, however, be related in some other form (other than straight line) e.g.,
quadratic (as we have seen in the above example), logarithmic or trigonometric form.
2. Some more properties of the correlation coefficient will be discussed in the next chapter on
Regression Analysis.
a×c
Property IV.
r(aX + b, cY + d) =
. r(X, Y)
…(8.8)
|a×c|
where | a × c | is the modulus value of a × c.
Remarks 1. Since correlation coefficient is independent of change of origin, we get :
a×c
r (aX + b, cY + d) = r(aX, cY) =
· r (X, Y).
|a×c|
2. Some Results on Covariance. If U = aX + b and V = CY + d, then
Cov (U, V) = ac . Cov (X, Y)
⇒
Cov [aX + b , CY + d ] = ac. Cov (X, Y)
…(i)
In particular, taking b = 0 = d, in (i) we get;
Cov (aX, cY) = ac. Cov (X, Y)
… (ii)
Taking a = c = 1 in (i), we get :
Cov (X + b, Y + d) = Cov (X, Y)
…(iii)
The above results can be restated as follows :
Cov (X + A, Y + B) = Cov (X, Y)
⎫
⎪
Cov (AX, BY) = AB . Cov (X, Y)
…(8.9)
⎬
⎪
⎭
Cov (AX + C, BY + D) = AB Cov (X, Y)
where A, B, C and D are constants.
Property V. If the variables x and y are connected by the linear equation ax + by + c = 0, then the
correlation coefficient between x and y is (+1) if the signs of a and b are different and (–1) if the signs of a
and b are alike.
8·13
CORRELATION ANALYSIS
Symbolically, if
ax + by + c = 0, then
+ 1‚ if a and b are of opposite signs
r = r (x, y) =
– 1‚ if a and b are of same sign
{
Example 8·10. The total of the multiplication of deviation of X and Y = 3044. No. of pairs of the
observations is 10. Total of deviations of X = – 170. Total of deviations of Y = – 20. Total of the squares of
deviations of X = 8288. Total of the squares of deviations of Y = 2264.
Find out the coefficient of correlation when the arbitrary means of X and Y are 82 and 68 respectively.
[Delhi Univ. B.Com. (Pass), 2001]
Solution. Let U = X – 82 and V = Y – 68. Then we are given :
n = 10, ∑UV = 3044,
∑U = – 170,
∑V = – 20,
∑U 2 = 8288,
∑V 2 = 2264.
Since the correlation coefficient (r) is independent of change of origin, we have
n ∑UV – (∑U) (∑V)
r(X, Y) = r (U, V) =
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
[n ∑U2 – (∑U)2] [n ∑V2 – (∑V)2]
=
=
10 × 3044 ± (±170)(±20)
2
2
[10 × 8288 ± (±170) ][10 × 2264 ± (±20) ]
27040
53980
22240
=
27040
232·34 × 149·13
=
30440 ± 3400
82880 ± 28900
22640 ± 400
27040
= 34648·86
= + 0·78
Example 8·11. Calculate correlation coefficient r(x, y) from the following data :
n = 10, ∑x = 140, ∑y = 150, ∑(x – 10) 2 = 180, ∑(y – 15) 2 = 215, ∑(x – 10) (y – 15) = 60
[Delhi Univ. B.Com. (Hons.), 2007]
Solution. Let us take u = x – 10 and v = y – 15. Then, we have
∑u = ∑(x – 10) = ∑x – 10 × n = 140 – 100 = 40
∑v = ∑(y – 15) = ∑y – 15 × n = 150 – 150 = 0
∑u2 = ∑(x – 10)2 = 180 ; ∑v2 = ∑(y – 15)2 = 215 ; ∑uv = ∑(x – 10) (y – 15) = 60
n ∑uv – (∑u) (∑v)
∴
rxy = ruv =
[n ∑u2 – (∑u)2 ] [n ∑v 2 – (∑v)2]
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
=
10 × 60 ± 40 × 0
[10 × 180 ± ( 40)2 ][10 × 215 ± 0]
=
600
200 × 2150
=
6
⎯⎯43
√
6
= 6·557 =
0·915.
Example 8·12. Calculate the coefficient of correlation for the ages of husbands and wives :
Age of Husband (years) 23,
27,
28
29,
30,
31,
33,
35,
36,
Age of Wife
(years) 18,
22,
23,
24,
25,
26,
28,
29,
30,
39
32
Solution.
CALCULATIONS FOR CORRELATION COEFFICIENT
x
23
27
28
29
30
31
33
35
36
39
y
18
22
23
24
25
26
28
29
30
32
u = x – 31
–8
–4
–3
–2
–1
0
2
4
5
8
v = y – 25
–7
–3
–2
–1
0
1
3
4
5
7
u2
64
16
9
4
1
0
4
16
25
64
v2
49
9
4
1
0
1
9
16
25
49
uv
56
12
6
2
0
0
6
16
25
56
∑x = 311
∑y = 257
∑u = 1
∑v = 7
∑u2 = 203
∑v2 = 163
∑uv = 179
8·14
BUSINESS STATISTICS
Karl Pearson’s correlation coefficient between U and V is given by
n∑uv – (∑u) (∑v)
10 × 179 ± 1 × 7
ruv =
=
[10 × 203 ± (1)2 ][10 × 163 ± (7)2 ]
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
[n∑u2 – (∑u)2 ] [n ∑v2 – (∑v)2]
1790 ± 7
=
(2030 ± 1)(1630 ± 49)
=
1783
2029 × 1581
1783
= 45·041783
× 39·76 = 1790·79 = 0·9956.
Since Karl Pearson’s correlation coefficient (r) is independent of change of origin, we get
rxy = ruv = 0·9956.
Note. It may be noted that the values of x and y, except for the last three pairs, are connected by the
linear relation : y = x – 5. Further, as the value of x decreases (increases), the value of y also decreases
(increases). Hence, we may expect a very high degree of positive correlation (almost perfect) with the value
of r approaching + 1.
Example 8·13. Find Karl Pearson’s coefficient of correlation between sales and expenses of the
following ten firms :
Firm
1
2
3
4
5
6
7
8
9
10
Sales (’000 units)
50
50
55
60
65
65
65
60
60
50
Expenses (’000 rupees)
11
13
14
16
16
15
15
14
13
13
Solution. Let sales (in thousand units) of a firm be denoted by X and expenses (in ’000 rupees) be
denoted by Y. It may be noted that we can take out factor 5 common in X series. Hence, it will be
convenient to change the scale also in X. Taking 65 and 13 as working means for X and Y respectively, let
us take :
u = (x – 65)/5 ; v = y – 13.
CALCULATIONS FOR CORRELATION COEFFICIENT
v = y – 13
u2
v2
uv
11
13
14
16
16
15
15
14
13
13
u = x – 65
5
–3
–3
–2
–1
0
0
0
–1
–1
–3
–2
0
1
3
3
2
2
1
0
0
9
9
4
1
0
0
0
1
1
9
4
0
1
9
9
4
4
1
0
0
6
0
–2
–3
0
0
0
–1
0
0
∑y =140
∑u = – 14
∑v = 10
∑u2 = 34
∑ v2 = 32
∑ uv = 0
Firms
x
y
1
2
3
4
5
6
7
8
9
10
50
50
55
60
65
65
65
60
60
50
∑x = 580
Karl Pearson’s correlation coefficient between u and v is given by
n∑uv – (∑u)(∑v)
10 × 0 ± (±14) × (10)
ruv =
=
[10 × 34 ± (±14)2 ] × [10 × 32 ± (10)2 ]
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
[n∑u2 – (∑u)2 ] [n∑v2 – (∑v)2]
=
140
(340 ± 196) × (320 ± 100)
=
140
144 × 220
=
140
31680
140
= 177·99
= 0·7866.
Since correlation coefficient is independent of change of origin and scale, we finally have :
rxy = ruv = 0·7866.
∑x
–x =
– ∑y 140
Aliter. We have :
= 580
10 = 58 ; y = n = 10 = 14.
n
Since x–and –y are integers, it will be convenient to compute r by taking the deviations from means
directly, i.e., by taking :
dx = x – –x = x – 58 ; dy = y – –y = y – 14, and use the formula (8·3a). [Try it.]
8·15
CORRELATION ANALYSIS
Example 8·14. Find Karl Pearson’s coefficient of correlation between the age and the playing habit of
the people from the following information. Also mention what does your calculated ‘r’ indicate.
Age group (years)
15 and less than 20
20 and less than 25
25 and less than 30
30 and less than 35
35 and less than 40
40 and less than 45
No. of people
200
270
340
360
400
300
No. of players
150
162
170
180
180
120
[Delhi Univ. B.Com. (Hons.), (External), 2006; Delhi Univ. B.Com. (Pass), 2002]
No. of
Percentage of players (y)
Solution. We want to find Karl Pearson’s Age group No. of
(yrs.)
people
players
(3)
correlation coefficient between the age and the
× 100
(4) =
(1)
(2)
(3)
(2)
playing habit of the people. To do this, we first
150
express the number of players in each age group
150
200
× 100 = 75
200
on a common base i.e., we find the number of 15—20
162
players out of a fixed number of persons (a 20—25
162
270
× 100 = 60
270
common base) which may be taken as 100 or
170
1000 or some other convenient figure. Here we 25—30
170
340
× 100 = 50
340
express the number of players as a percentage
180
of the total people in each age group.
180
360
30—35
× 100 = 50
360
Now we compute Karl Pearson’s
180
180
400
× 100 = 45
correlation coefficient between age (x) and the 35—40
400
percentage of players in each age group (y).
120
Age-group
Mid-value
(x)
15—20
20—25
25—30
30—35
35—40
40—45
17·5
22·5
27·5
32·5
37·5
42·5
120
300
40—45
300
CALCULATIONS FOR CORRELATION COEFFICIENT
y – 50
x – 27·5
v=
u=
y
u2
5
5
75
60
50
50
45
40
Total
× 100 = 40
v2
uv
–2
–1
0
1
2
3
5
2
0
0
–1
–2
4
1
0
1
4
9
25
4
0
0
1
4
– 10
–2
0
0
–2
–6
∑u = 3
∑v = 4
∑u2 = 19
∑v2 = 34
∑uv = – 20
Since correlation coefficient is independent of change of origin and scale we have :
n∑uv – (∑u)(∑v)
6 × (±20) ± (3) × ( 4)
rxy = ruv =
=
2
2
2
2
[6 × 19 ± (3)2 ].[6 × 34 ± ( 4)2 ]
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
[n∑u – (∑u) ] . [n∑v – (∑v) ]
=
±120 ± 12
(114 ± 9) × (204 ± 16)
=
– 132
⎯⎯⎯⎯⎯⎯⎯
√
105 × 118
=
– 132
⎯⎯⎯⎯⎯
√
19740
– 132
= 140·4991 =
– 0·9395.
Thus, we conclude that there is a very high degree of negative correlation, (almost perfect negative
correlation) between age (x) and playing habit (y). This implies that with advancement in age, the people’s
interest in playing goes on decreasing and the scatter diagram of the (x, y) values gives points clustering
almost around a straight line starting from left top and going to right bottom.
Example 8·15. (i) Compute the correlation coefficient between the corresponding values of X and Y in
the following table :
X
Y
2
18
4
12
5
10
6
8
8
7
11
5
8·16
BUSINESS STATISTICS
(ii) Multiply each X value in the table by 2 and add 6. Multiply each value of Y in the table by 3 and
subtract 15. Find the correlation coefficient between the two new sets of values. Explain why you do or do
not obtain the same result as in (i).
Solution. (i)
COMPUTATION OF CORRELATION COEFFICIENT
—
—
X–X=X–6
X
Y
2
4
5
6
8
11
18
12
10
8
7
5
∑X =36
∑Y = 60
–4
–2
–1
0
2
5
8
2
0
–2
–3
–5
—
X =
1
6 ∑X
∑ (Y – Y ) = 0
= 36
6 = 6,
—
and
—
(Y – Y )2
16
4
1
0
4
25
—
∑(X – X ) = 0
—
We have :
—
(X – X )2
Y – Y = Y – 10
64
4
0
4
9
25
—
—
(X – X ) (Y – Y )
– 32
–4
0
0
–6
– 25
—
—
—
∑(X – X )2 = 50 ∑(Y – Y )2 = 106 ∑(X – X ) (Y – Y ) = – 67
—
Y = 61 ∑Y = 60
6 = 10
Hence the correlation coefficient between X and Y is given by :
—
—
∑(X – X ) (Y – Y )
– 67
– 67
– 67
rxy =
=
= 72·81 = – 0·92
— 2
— 2 1/2 =
⎯⎯⎯⎯⎯⎯
√
50 × 106 √
⎯⎯⎯⎯
5300
[∑(X – X ) ∑(Y – Y ) ]
Hence, the variables X and Y are highly negatively correlated.
(ii) Let us define new variables U and V as :
U = 2X + 6 and V = 3Y – 15
We are now required to find the correlation coefficient between the new sets of values of variables U
and V as given in the following table.
COMPUTATION OF CORRELATION COEFFICIENT BETWEEN U AND V
X
2
4
5
6
8
11
Y
18
12
10
8
7
5
Totals
∴
ruv =
=
U = 2X + 6
10
14
16
18
22
28
V = 3Y – 15
39
21
15
9
6
0
U2
100
196
256
324
484
784
V2
1521
441
225
81
36
0
∑U = 108
∑V = 90
∑U 2 = 2144
∑V 2 = 2304
n∑UV – (∑U) (∑V)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
n∑U2 – (∑U)2 √
⎯⎯⎯⎯⎯⎯⎯⎯⎯
n∑V2 – (∑V) 2
7308 ± 9720
(12864 ± 11664)(13824 ± 8100)
=
=
UV
390
294
240
162
132
0
∑UV = 1218
6 × 1218 ± 108 × 90
6 × 2144 ± (108)2 × 6 × 2304 ± (90)2
±2412
1200 × 5724
=
– 2412
⎯⎯⎯⎯⎯⎯
√
6868800
– 2412
= 2620·8395 =
– 0·92
Example 8·16. If the relation between two random variables x and y is 2x + 3y = 4, then the
correlation coefficient between them is :
(i) – 2/3, (ii) 1, (iii) – 1, (iv) none of these.
[I.C.W.A. (Intermediate), June 2000]
Solution. Since x and y are connected by the linear relation :
2x + 3y = 4
⇒
y = – 32 x + 43 ,
… (*)
there is perfect correlation between x and y, i.e., r = ± 1. Further from (*), we observe that as x increases, y
decreases. Hence, there is perfect negative correlation between x and y.
∴
r=–1
⇒
(iii) is the correct answer.
CORRELATION ANALYSIS
8·17
Aliter. If x and y are connected by the linear relation ax + by + c = 0, then
+ 1‚ if a and b have opposite signs.
r = r (x, y) =
– 1‚ if a and b have same sign.
{
We are given 2x + 3y = 4. Since a = 2 and b =3, have the same sign, r = r(x, y) = – 1.
Example 8·17. For the bivariate data [(x, y)] = [(20, 5), (21, 4), (22, 3)], the correlation coefficient
between x and y is :
(i) 0, (ii) 1, (iii) – 1, (iv) 0·5.
[I.C.W.A. (Intermediate), June 2002]
Solution. From the given data, we observe that
20 + 5 = 25, 21 + 4 = 25 and 22 + 3 = 25.
Thus, x and y are connected by the linear relation : x + y = 25
… (*)
⇒
There is perfect correlation between x and y
⇒ r=±1
… (**)
From (*), we get
y = 25 – x
∴ As x increases, y decreases (by the same amount).
⇒
x and y are negatively correlated
… (***)
From (**) and (***), we conclude that r = r (X, Y) = – 1.
Aliter. We can compute the value of r (X, Y) from the given data by using the formula. This is left as
an exercise to the reader.
Example 8·18. (a) The correlation coefficient between two variables X and Y is found to be 0·4. What
is the correlation between 2x and (– y) ?
[Delhi Univ. B.A. (Econ., Hons.), 1997]
(b) “If the correlation coefficient between two variables x and y is positive, then the coefficient of
correlation between – x and – y is also positive” Comment.
[Delhi Univ. B.A. (Econ. Hons.), 1996]
Solution. (a) We are given : r(x, y) = 0·4
… (i)
a×c
We know that :
r (aX, cY) =
· r (X, Y)
… (*)
|a|× |c|
Using (*) and (i), we get :
2 × (– 1)
– 2 × 0·4
r (2x, – y) = r (2x, – 1·y) =
r (x, y) =
= – 0·4
|2|×|–1|
2×1
(b) We are given : r(x, y) > 0… (ii)
Using (*), we get
r (– x , – y) = r (– 1.x , – 1.y) = (– 1) × (– 1) . r(x, y) = r(x, y)
|–1|×|–1|
Hence, if r(x, y) is positive, then r (– x, – y) is also positive.
8·4·2. Assumptions Underlying Karl Pearson’s Correlation Coefficient. Pearsonian correlation
coefficient r is based on the following assumptions :
(i) The variables X and Y under study are linearly related. In other words, the scatter diagram of the
data will given a straight line curve.
(ii) Each of the variables (series) is being affected by a large number of independent contributory
causes of such a nature as to produce normal distribution. For example, the variables (series) relating to
ages, heights, weights, supply, price, etc., conform to this assumption. In the words of Karl Pearson :
“The sizes of the complex organs (something measureable) are determined by a great variety of
independent contributing causes, for example, climate, nourishment, physical training and innumerable
other causes which cannot be individually observed or their effects measured.” Karl Pearson further
observes, “The variations in intensity of the contributory causes are small as compared with their absolute
intensity and these variations follow the normal law of distribution.”
(iii) The forces so operating on each of the variable series are not independent of each other but are
related in a causal fashion. In other words, cause and effect relationship exists between different forces
operating on the items of the two variable series. These forces must be common to both the series. If the
8·18
BUSINESS STATISTICS
operating forces are entirely independent of each other and not related in any fashion, then there cannot be
any correlation between the variables under study.
For example the correlation coefficient between :
(a) the series of heights and income of individuals over a period of time,
(b) the series of marriage rate and the rate of agricultural production in a country over a period of time,
(c) the series relating to the size of the shoe and intelligence of a group of individuals, should be zero,
since the forces affecting the two variable series in each of the above cases are entirely independent of each
other.
However, if in any of the above cases the value of r for a given set of data is not zero, then such
correlation is termed as chance correlation or spurious or non-sense correlation.
[Also see § 8·1·2.]
8·4·3. Interpretation of r. The following general points may be borne in mind while interpreting an
observed value of correlation coefficient r :
(i) r = + 1 implies that there is perfect positive correlation between the variables. In other words, the
scatter diagram will be a straight line starting from left bottom and rising upwards to the right top as shown
in Fig. 8·1, § 8·3.
(ii) If r = – 1, there is perfect negative correlation between the variables. In this case scatter diagram
will again be a straight line as shown in Fig. 8·1, § 8·3.
(iii) If r = 0, the variables are uncorrelated. In other words, there is no linear (straight line) relationship
between the variables. However, r = 0 does not imply that the variables are independent [c.f. Property III,
page 8·11].
(iv) For other values of r lying between + 1 and – 1, there are no set guidelines for its interpretation.
The maximum we can conclude is that nearer is the value of r to 1, the closer is the relation between the
variables and nearer is the value of r to 0, the less close is the relationship between them. One should be
very careful in interpreting the value of r as it is often mis-interpreted.
(v) The reliability or the significance of the value of the correlation coefficient depends on a number of
factors. One of the ways of testing the significance of r is finding its probable error [c.f. § 8·5], which in
addition to the value of r takes into account the size of the sample also.
(vi) Another more useful measure for interpreting the value of r is the coefficient of determination [c.f.
§ 8·9]. It is observed there that the closeness of the relationship between two variables is not proportional
to r.
8·5. PROBABLE ERROR
After computing the value of the correlation coefficient, the next step is to find the extent to which it is
dependable. Probable error of correlation coefficient, usually denoted by P.E.(r) is an old measure of
testing the reliability of an observed value of correlation coefficient in so far as it depends upon the
conditions of random sampling.
If r is the observed correlation coefficient in a sample of n pairs of observations then its standard error,
usually denoted by S.E. (r) is given by :
1 – r2
S.E.(r) =
… (8·10)
⎯n
√
Probable error of the correlation coefficient is given by :
(1 – r2 )
P.E. (r) = 0·6745 × S.E. (r) = 0·6745
… (8·11)
⎯n
√
The reason for taking the factor 0·6745 is that in a normal distribution 50% of the observations lie in
the range μ ± 0·6745 σ, where μ is the mean and σ is the s.d.
According to Secrist, “The probable error of the correlation coefficient is an amount which if added to
and subtracted from the mean correlation coefficient, produces amounts within which the chances are even
that a coefficient of correlation from a series selected at random will fall.”
8·19
CORRELATION ANALYSIS
Uses of Probable Error
1. The probable error of correlation coefficient may be used to determine the limits within which the
population correlation coefficient may be expected to lie.
Limits for population correlation coefficient are
r ± P.E. (r)
… (8·12)
This implies that if we take another random sample of the same size n from the same population from
which the first sample was taken, then the observed value of the correlation coefficient, say, r1 in the
second sample can be expected to lie within the limits given in (8·12).
2. P.E. (r) may be used to test if an observed value of sample correlation coefficient is significant of
any correlation in the population. The following guidelines may be used :
(i) If r < P.E. (r) i.e., if the observed value of r is less than its P.E., then correlation is not at all
significant.
(ii) If r > 6 P.E. (r) i.e., if observed value of r is greater than 6 times its P.E., then r is definitely
significant.
(iii) In other situations, nothing can be concluded with certainty.
Important Remarks 1. Sometimes, P.E. may lead to fallacious conclusions particularly when n, the
number of pairs of observations, is small. In order to use P.E. effectively, n should be fairly large.
However, a rigorous test for testing the significance of an observed sample correlation coefficient is
provided by Student’s t test.
2. P.E. can be used only under the following conditions :
(i) The data must have been drawn from a normal population.
(ii) The conditions of random sampling should prevail in selecting sampled observations.
Example 8·19. (a) Find Karl Pearson’s coefficient of correlation from the following series of marks
secured by 10 students in a class test in Mathematics and Statistics :
Marks in Mathematics.
:
45
70
65
30
90
40
50
75
85
60
Marks in Statistics
:
35
90
70
40
95
40
60
80
80
50
Also calculate its probable error. Assume 60 and 65 as working means.
(b) Hence discuss if the value of r is significant or not. Also compute the limits within which the
population correlation coefficient may be expected to lie.
Solution. (a) Let the marks in mathematics be denoted by the variable X and the marks in statistics by
the variable Y. It may be noted that we can take out the factor 5 common in each of the X and Y series.
Hence, it will be convenient to change the scale also. Taking 60 and 65 as working means for X and Y series
respectively, let us take :
x – 60
y – 65
and
v=
u=
5
5
CALCULATIONS FOR CORRELATION COEFFICIENT
x
y
u
v
u2
v2
uv
18
36
9
–6
–3
35
45
n∑uv – (∑u)(∑v)
r =
10
25
4
5
2
90
70
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
[n∑u2 – (∑u)2 ] × [n∑v 2 – (∑v)2]
1
1
1
1
1
70
65
10 × 141 ± 2 × (±2)
=
30
25
36
–5
–6
40
30
(10 × 140 ± 4)(10 × 176 ± 4)
36
36
36
6
6
95
90
1414
=
20
25
16
–5
–
4
40
40
1396 × 1756
2
1
4
–1
–2
60
50
⇒ log r = log 1414 – 21 [log 1396 + log 1756]
9
9
9
3
3
80
75
= 3·1504 – 12 [3·1449 + 3·2445]
15
9
25
3
5
80
85
0
9
0
–3
0
50
60
= 3·1504 – 12 × 6·3894
We have :
–
= 3·1504 – 3·1947 = – 0·0443 = 1·9557
Total :
2
–2
140
176
141
8·20
⇒
∴
BUSINESS STATISTICS
–
r = Antilog ( 1·9557) = 0·9031 = 0·9
rxy = ruv ~– 0·9.
Probable Error of Correlation Coefficient is given by :
P.E. (r) = 0·6745
1 – r2
⎯
√n
=
0·6745 × 0·19
⎯⎯10
√
=
0·128155
3·1623
= 0·0405.
(b) Significance of r. We have
r = 0·9 and 6 P.E. (r) = 6 × 0·0405 = 0·2430.
Since r is much greater than 6 P.E. (r), the value of r is highly significant.
Remark. Since the value of r is significant, it implies that ordinarily, higher the marks of a candidate
in Mathematics, higher is his score in Statistics also and lower the marks of a candidate in Mathematics,
lower is his score in Statistics also. However, it does not mean that all the students who are good in
Mathematics are also good in Statistics and all those students who are poor in Mathematics are also poor in
Statistics. It should be clearly borne in mind that “the coefficient of correlation expresses the relationship
between two series and not between the individual items of the series”.
Limits for Population Correlation Coefficient are :
r ± P.E.(r) = 0·9031 ± 0·0405 i.e., 0·8626 and 0·9436.
This implies that if we take another sample of size 10 from the same population, then its correlation
coefficient can be expected to lie between 0·8626 and 0·9436.
Example 8·20. Test the significance of correlation for the following values based on the number of
observations (i) 10, and (ii) 100, r = + 0·4 and + 0·9.
Solution. We know that an observed value of r is definitely significant if
r
>6
r > 6 P.E.(r)
⇒
P.E.(r)
In this case, we have :
No. of observations
r
P.E. (r)
10
0·4
0·6745
100
0·4
0·6745
10
0·9
0·6745
100
0·9
0·6745
1 – (0·4)2
⎯⎯10
√
1 – (0·4)2
⎯⎯⎯
√
100
1 – (0·9)2
⎯⎯10
√
1 – (0·9)2
⎯⎯⎯
√
100
r
P.E. (r)
Significance of r.
= 0·18
0·4
= 2·22 < 6
0·18
Not significant
= 0·06
0·4
= 6·67 > 6
0·06
Significant
= 0·04
0·9
= 22·5 > 6
0·04
Highly significant
= 0·128
0·9
=7>6
0·13
Significant
EXERCISE 8·2
1. Explain the meaning and significance of the concept of correlation. How will you calculate it from statistical
point of view.
2. (a) Define Karl Pearson’s coefficient of correlation. What is it intended to measure ?
(b) What are the special characteristics of Karl Pearson’s coefficient of correlation ? What are the underlying
assumptions on which this formula is based ?
(c) How do you interpret a calculated value of Karl Pearson’s coefficient of correlation ? Discuss in particular the
values of r = 0, r = – 1 and r = + 1.
3. (a) Explain what is meant by coefficient of correlation between two variables. What are the different methods of
finding correlation ? Distinguish between Positive and Negative correlation.
[Calicut Univ. B.Com., 1997]
(b) Write down an expression for the Karl Pearson’s coefficient of linear correlation. Why is it termed as the
coefficient of linear correlation ? Explain.
[Delhi Univ. B.A. (Econ., Hons.), 1997]
4. Define product moment correlation coefficient between two variables x and y. State its limits. Draw the scatter
diagram for the extreme cases.
8·21
CORRELATION ANALYSIS
5. (a) If x and y are independent variates then prove that they are uncorrelated. Is the converse true ? Explain your
answer with the help of an example.
(b) Prove that two independent variables are uncorrelated. By giving an example, show that the converse is not
true. Explain the reason ?
[Guru Nanak Dev Univ. MBA, 1994]
(c) Comment on the following statement :
If the coefficient of correlation between two variables is zero, it does not mean that the variables are unrelated.
[Delhi Univ. B.Com. (Hons.), 2002]
6. Discuss the statistical validity of the following statements :
(a) “High positive coefficient of correlation between increase in the sale of a newspaper and increase in the
number of crimes, leads to the conclusion that newspaper reading may be responsible for the increase in the number of
crimes.”
(b) “A high positive value of r between the increase in cigarette smoking and increase in lung cancer establishes
that cigarette smoking is responsible for lung cancer.”
(c) If the coefficient of correlation between the annual value of exports during the last ten years and the annual
number of children born during the same period is + 0·9, what inference, if any, would you draw ?
[Delhi Univ. B.A. (Econ. Hons.), 1996]
7. Comment on the following :
(a) “Positive correlation r = 0·9, is found between the number of children born and exports over last decade.”.
[Delhi Univ. B.Com. (Hons.), 2001]
(b) The correlation coefficient between the railway accidents in a particular year and the babies born in that year
was found to be 0·8.
8. (a) Define a scatter diagram. Draw the scatter diagram when (i) r = + 1, (ii) r = – 1 and (iii) r = 0, where r is the
correlation coefficient.
[I.C.W.A. (Intermediate), Dec. 2001]
(b) What is a scatter diagram ? Give the procedure of drawing a scatter diagram. Draw scatter diagrams when the
coefficient of correlation r = + 1 and r = – 1.
[C.A. (Foundation), May 2000]
9. The production manager of a company maintains that the flow time in days (y), depends on the number of
operations (x) to be performed. The following data give the necessary information :
x :
2
2
3
4
4
5
6
6
7
7
y :
8
13
14
11
20
10
22
26
22
25
Plot a scatter diagram. Calculate the value of the Karl Pearson’s Product Moment Correlation Coefficient .
[I.C.W.A. (Intermediate), Dec. 1995]
Ans. r (x, y) = 0·78.
10. Making use of the data given below, calculate the coefficient of correlation r12
Case :
A
B
C
D
E
F
G
H
X1
:
10
6
9
10
12
13
11
9
X2
:
9
4
6
9
11
13
8
4
Ans. r12 = 0·8958.
11. Calculate Karl Pearson’s coefficient of correlation from the following data, using 20 as the working mean for
price and 70 as the working mean for demand :
Price
:
14
16
17
18
19
20
21
22
23
Demand
:
84
78
70
75
66
67
62
58
60
[Delhi Univ. B.Com. (Pass), 1999]
Ans. r = – 0·954.
12. Calculate the Karl Pearson’s coefficient of correlation from the following data :
Percentage of Marks
Percentage of Marks
No.
1.
2.
3.
4.
Subject
Hindi
English
Economics
Accounts
Ans. r = 0·623.
First Term
75
81
70
76
Second Term
62
68
65
60
No.
5.
6.
7.
8.
Subject
First Term Second Term
Commerce
77
69
Mathematics
81
72
Statistics
84
76
Costing
75
72
[Delhi Univ. B.Com. (Pass), 2000]
8·22
BUSINESS STATISTICS
13. Calculate the Karl Pearson’s coefficient of correlation for the following ages of husbands and wives at the time
of their marriage :
Age of husband (in years)
:
23
27
28
28
28
30
30
33
35
38
Age of wife (in years)
:
18
20
22
27
21
29
27
29
28
29
Ans. r = 0·8013.
14. Calculate the Pearson’s coefficient of correlation from the following data using 44 and 26 respectively as the
origin of X and Y :
X:
43
44
46
40
44
42
45
42
38
40
42
57
Y:
29
31
19
18
19
27
27
29
41
30
26
10
[Osmania Univ. B.Com., 1998]
Ans. rxy = – 0·7326.
15. The following table gives the distribution of the total population and those who are totally or partially blind
among them. Find out if there is any relation between age and blindness.
Age (Years)
:
0—10
10—20 20—30 30—40 40—50 50—60 60—70 70—80
No. of Persons (’000)
:
100
60
40
36
24
11
6
3
Blind
:
55
40
40
40
36
22
18
15
Hint. Here we shall find the correlation coefficient between age (X) and the number of blinds per lakh (Y) as
given in the following table.
X
5
15
25
35
45
55
65
75
Y
55
67
100
111
150
200
300
500
Ans. r = 0·8982.
16. With the following data in 6 cities, calculate the coefficient of correlation by Pearson’s method between the
density of population and the death rate.
Cities
Area in square miles
Population (in ’000)
No. of deaths
A
150
30
300
B
180
90
1440
C
100
40
560
D
60
42
840
E
120
72
1224
F
80
24
312
[C.A. (Intermediate). May 1981]
Population
No. of deaths
Hint. Find r between, Density =
; and Death Rate =
× 1000.
Area
Population
Ans. r = 0·9876.
17. Calculate the correlation coefficient from the following data :
X:
12
9
8
10
11
13
7
Y:
14
8
6
9
11
12
3
Let now each value of X be multiplied by 2 and then 6 be added to it. Similarly multiply each value of Y by 3 and
subtract 2 from it. What will be the correlation coefficient between the new series of X and Y.
[C.A. (Foundation), May 1997]
Ans. Let U = 2X + 6, V = 3Y – 2. Since correlation coefficient is independent of change of origin and scale,
r(U, V) = r(X, Y) = 0·9485.
18. (a) Given : ∑X = 125,
∑Y = 100,
∑X 2 = 650,
∑Y2 = 436,
∑XY = 520
and
n = 25,
obtain the value of Karl Pearson’s correlation coefficient r(X, Y).
Ans. 0·67.
19. You are given the following information relating to a frequency distribution comprising of 10 observations.
—
—
X = 5·5,
Y = 4·0,
∑X 2 = 385,
∑Y 2 = 192;
Find rxy.
Hint. Use ∑(X + Y) 2 = ∑X 2 + ∑Y 2 + 2 ∑XY and find ∑XY = 185.
Ans. r(X, Y) = – 0·681.
∑(X + Y) 2 = 947.
[Punjab Univ. B.Com., 1994]
8·23
CORRELATION ANALYSIS
20. A computer while calculating the correlation coefficient between the variables X and Y obtained the following
results :
N = 30, ∑X = 120, ∑X 2 = 600, ∑Y = 90, ∑Y2 = 250, ∑XY = 356
It was, however, later discovered at the time of checking that it had copied down two pairs of observations as :
X
Y
X
Y
,
while the correct values were,
8
10
8
12
12
7
10
8
Obtain the correct value of the correlation coefficient between X and Y.
[I.C.W.A. Dec., 2003]
Ans. r = 0·0504
21. Coefficient of correlation between X and Y for 20 items is 0·3; mean of X is 15 and that of Y 20, standard
deviations are 4 and 5 respectively. At the time of calculations one pair (x = 27, y = 30) was wrongly taken as
(x = 17, y = 35). Find the correct coefficient of correlation.
[Delhi Univ. B.Com. (Hons.), (External), 2007]
Ans. Correct value of correlation coefficient = 0·5153.
22. In order to find the correlation coefficient between two variables X and Y from 12 pairs of observations, the
following calculations were made :
∑X = 30, ∑Y = 5, ∑X2 = 670, ∑Y2 = 285, ∑XY = 334
On subsequent verification it was found that the pair (X = 11, Y = 4) was copied wrongly, the correct value being
(X = 10, Y = 14). Find the correct value of correlation coefficient.
Ans. 0·78.
23. What do you understand by the probable error of correlation coefficient ? Explain how it can be used to :
(i) Interpret the significance of an observed value of sample correlation coefficient.
(ii) Determine the limits for the population correlation coefficient.
24. Calculate the coefficient of correlation and find its probable error from the following data :
X:
7
6
5
4
3
2
1
Y:
18
16
14
12
10
6
8
Ans. rxy = 0·9643; P.E. (r) = 0·0179.
25. Find Karl Pearson’s correlation coefficient between age and playing habit of the following students :
Age (years)
:
15
16
17
18
19
20
No. of students
:
250
200
150
120
100
80
Regular players
:
200
150
90
48
30
12
Also calculate the probable error and point out if coefficient of correlation is significant.
[Himachal Pradesh Univ. M.B.A. 1998; Delhi Univ. B.Com. (Hons.), 1996]
Hint. Find r between age (X) and percentage of regular players (Y).
Ans. rxy = – 0·9912 ; P.E.(r) = 0·0048; r is highly significant.
26. Calculate Karl Pearson’s coefficient of correlation for the following series.
Price (in Rs.)
:
110—111
111—112
112—113
113—114
Demand (in kg.)
:
600
640
640
680
Price (in Rs.)
:
116—117
117—118
118—119
Demand (in kg.)
:
830
900
1,000
114—115
700
115—116
780
Also calculate the probable error of the correlation coefficient. From your result can you assert that the demand is
correlated with price ?
Hint. Find correlation coefficient between x : Mid-value of price (in Rs.) and y demand (in kg.)
Ans. r = 0·9651 ; P.E. (r) = 0·0154.
27. (a) A student calculates the value of r as 0·7 when the number of items (n) in the sample is 25. Find the limits
within which r lies for another sample from the same universe.
Ans. Required limits for r are 0·767 and 0·633.
(b) A student calculates the value of r as 0·7 when the value of N is 5 and concludes that r is highly significant. Is
he correct ?
[Delhi Univ. B.Com. (Hons.), 1997]
Ans.
r
0·7 ⎯
√ 5 = 4·55 < 6. Not significant.
=
P.E.(r) 0·6745 × 0·51
8·24
BUSINESS STATISTICS
28. The correlation coefficient between Physics and Mathematics final marks for a group of 21 students was
computed to be 0·80. Find 95% confidence limits for the coefficient.
Ans. Required limits = r ± 1·96 P.E.(r) = 0·8 ± 1·96 × 0·05299 = 0·6961 and 0·9039.
29. The deviations from the respective means of X and Y series are given below :
x:
–4
–3
–2
–1
0
1
2
y:
3
–3
–4
0
4
1
2
3
–2
4
–1
Calculate Karl Pearson’s coefficient of correlation from the above data.
[Delhi Univ. B.Com. (Pass), 1995]
1
Hint. Cov (X, Y) = ∑xy = 0.
n
Ans. r(X,Y) = 0.
30. Calculate the coefficient correlation between X and Y series from the following data :
X series
Y series
No. of observations
15
15
Arithmetic mean
25
18
Standard deviation
5
5
∑[(X – 25) (Y – 18)] = 125.
—
[Delhi Univ. B.Com. (Pass), 1995]
—
∑(X – X ) (Y – Y) ∑(X – 25) (Y – 18)
125
=
=
= 0·33
15 × 5 × 5
15 × 25
n σX σY
31. Given n = 10;
∑x = 100;
∑y = 150; ∑(x – 10) 2 = 180; ∑(y – 15) 2 = 25;
∑(x – 10) (y – 15) = 60;
Obtain Karl Pearson’s correlation coefficient.
[Ans. (2/√
⎯ 5) = 0·8944]
32. In a correlation study the results were
—
—
∑xy = 40,
N = 100,
∑x2 = 80,
∑y2 = 20, where x = X – X and y = Y – Y
The correlation coefficient is :
(a) + 1·0
(b) –1·0
(c) zero
(d) None of these.
Ans. (a).
Ans. rXY =
33. Given r = 0·8,
∑xy = 60,
σy = 2·5 and ∑x2 = 90
Find the number of items. Here x and y are deviations from respective means.
Ans. n = 10.
34. (a) The following results are obtained between two series. Compute the coefficient of correlation.
X Series
Number of items
7
Arithmetic mean
4
Sum of square of deviations from arithmetic mean
28
Summation of products of deviations of X and Y series from their respective means = 46
Y Series
7
8
76
Ans. r = 0·997.
(b) From the following data calculate the coefficient of correlation between two variables X and Y :
(i) Number of items in X-series or Y-series = 12.
(ii) Sum of the squares of deviation from mean : 360 for X-series and 250 for Y-series.
(iii) Sum of the products of deviations of the two series from their respective means = 225.
Ans. rxy = 0·75.
35. (a) The coefficient of correlation between two variables X and Y is 0·38. Their covariance is 10·2. The variance
of X is 16. Find the standard deviation of Y-series.
Ans. σy = 6·71.
(b) The covariance between the length and weight of five items is 6 and their standard deviations are 2·45 and 2·61
respectively. Find the coefficient of correlation between length and weight.
[Delhi Univ. B.Com. (Hons.), 2000]
Ans. r = 0·9383.
36. (a) The coefficient of correlation between two variates X and Y is 0·8 and their covariance is 20. If the variance
of X series is 16, find the standard deviation of Y series.
[C.A. (Foundation), June 1993]
Ans. σy = 6·25.
CORRELATION ANALYSIS
8·25
(b) The coefficient of correlation between two variables X and Y is 0·4 and their co-variance is 10. If variance of X
series is 9, find the second moment about mean of Y series.
[Delhi Univ. B.Com. (Hons.), 1996]
Hint. Second moment about mean of Y series is σy 2.
Ans. σy2 = (8·33)2 = 69·39.
37. (a) For the bivariate data : {(x, y) = (10, 4), (11, 3), (12, 2) (14, 0), (8, 6)}, the coefficient of correlation
between x and y is :
(i) – 1
(ii) 0·5
(iii) 1
(iv) 0
(b) For the bivariate data : {(x, y) : (15, 3), (20, 8), (25, 13), (30, 18)} the coefficient of correlation between x and y
is :
(i) 1
(ii) – 1
(iii) 0
(iv) 0·5
Ans. (a) x + y = 14 ; r = – 1
(b) x – y = 12 ; r = + 1.
38. (a) Given that the correlation between x and y is 0·5, what is the correlation between 2x – 4 and 3 – 2y ?
Ans. r = – 0·5.
(b) Point out the inconsistency in the statement :
“The correlation coefficient of 3x and – 2y is the same as the correlation coefficient of x and y.”
[I.C.W.A. (Intermediate), Dec. 1998]
Ans. r(3x, – 2y) = – r(x, y). Statement is inconsistent.
39. (a) The corresponding values of the variables are given below :
X:
2
3
5
8
9
Y:
4
6
10
16
18
The correlation coefficient between the variables is : – 1, 0, 1 or none of these. Justify your answer.
Ans. rxy = + 1.
(b) Let r be the correlation between x and y. What is the correlation coefficient between :
(i) (3x + 1) and (2y – 3), and (ii) x and – y ?
Explain your answers.
Ans. (i) r, (ii) – r.
(c) If U = 2x + 11 and V = 3y + 7, what will be correlation coefficient between U and V ? Justify your statement.
Ans. ruv = rxy.
(d) Are the following statements valid ? Justify your answer :
(i) Positive value of correlation coefficient between x and y implies that if x decreases, y tends to increase.
(ii) Correlation coefficient is independent of the origin of reference but is dependent on the units of
measurement.
(iii) Correlation coefficient between x and y turned out to be 1·25.
Ans. (i) False, (ii) False, (iii) Impossible.
40. Comment on the following, giving reasons for your conclusions :
(a) If the correlation coefficient between two variables X and Y is positive, then
(i) the correlation coefficient between – X and – Y is positive.
(ii) the correlation coefficient between X and – Y or – X and Y is positive.
(b) The correlation coefficient between two variables is 1·4.
(c) If the variables are independent then they are uncorrelated.
(d) Correlation coefficient can be calculated only if the two variables are measured in the same units.
(e) If the correlation coefficient between two variables is zero, then the variables are independent.
(f) The value of r cannot be negative.
(g) r measures every type of relationship between the two variables.
(h) “The closeness of relationship between two variables is proportional to r.”
(i) A student while studying correlation between smoking and drinking found a value of r as high as 1·62.
8·6. CORRELATION IN BIVARIATE FREQUENCY TABLE
If in a bivariate distribution the data are fairly large, they may be summarised in the form of a two-way
table. Here for each variable, the values are grouped into various classes (not necessarily the same for both
the variables), keeping in view the same considerations as in the case of univariate distribution. For
8·26
BUSINESS STATISTICS
example, if there are m classes for the X-variable series and n classes for the Y-variable series then there
will be m × n cells in the two-way table. By going through the different pairs of the values (x, y) and using
tally marks we can find the frequency for each cell and thus obtain the so-called bivariate frequency table
as shown below.
BIVARIATE FREQUENCY TABLE
X Series
Y Series
↓
Classes
→
x1
x2
...
Mid Points
...
x
...
Total of
frequencies
of Y
xm
Mid Points
Classes
y1
y2
fy
f (x, y)
y
..
.
yn
Total of
frequencies
of X
Total
∑ fx = ∑ fy = N
fx
Here f (x, y) is the frequency of the pair (x, y).
The formula for computing the correlation coefficient between X and Y for the bivariate frequency
table is
N ∑ x y f (x‚ y) – (∑ x fx) (∑ y fy)
r =
… (8·13)
[N ∑ x 2 fx – (∑ x fx)2] × [N ∑ y2 fy – (∑ y fy)2]
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
where N is the total frequency. If there is no confusion we may use the formula :
N∑ f x y – (∑ f x) (∑ f y)
r = rxy =
… (8·13a)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
[N ∑ f x2 – (∑ f x)2] × [N ∑ f y2 – (∑ f y)2]
where the frequency f used for the product xy is nothing but f(x, y) and the frequency f used in the sums ∑fx
and ∑fy are respectively the frequencies of x and y, viz., fx & fy as explained in the above table. If we change
the origin and scale in X and Y by transforming them to the new variables U and V by
x–A
y–B
u=
and
v=
; h > 0, k > 0
h
k
where h and k are the widths of the x-classes and y-classes respectively and A and B are constants, then by
Property II of r, we have :
N∑ f u v – (∑ f u) (∑ f v)
rxy = ruv =
… (8·14)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
[N ∑ f u2 – (∑ f u) 2 ] × [N ∑ f v2 – (∑ f v)2]
We shall explain the method by means of examples.
Example 8·21. Family income and its percentage spent on food in the case of hundred families gave
the following bivariate frequency distribution. Calculate the coefficient of correlation and interpret its
value.
Food Expenditure
(in %)
10—15
15—20
20—25
25—30
200—300
—
—
7
3
300—400
—
4
6
10
Family income (Rs.)
400—500
500—600
—
3
9
4
12
5
19
8
600—700
7
3
—
—
[Delhi Univ. M.B.A., 2000]
8·27
CORRELATION ANALYSIS
Solution. Let us denote the income (in Rupees) by the variable X and the food expenditure (%) by the
variable Y.
Steps 1. Find the mid-points of various classes for X and Y series.
2. Change the origin and scale in X-series and Y-series by transforming them to new variables u and v
as defined below :
y – B y – 17·5
x – A x – 450
u=
= 100 ,
and
v=
= 5 ,
h
k
where x denotes the mid-points of the X-sereis and y denotes the mid-points of the Y-series, and h and k are
the magnitudes of the classes of X and Y series respectively.
3. For each class of X, find the total of cell frequencies of all the classes of Y and similarly for each
class of Y, find the total of cell frequencies of all the classes of X.
4. Multiply the frequencies of x by the corresponding values of the variable u and find the sum ∑fu .
5. Multiply the frequencies of y by the corresponding values of the variable v and find the sum ∑fv.
6. Multiply the frequency of each cell by the corresponding values of u and v and write the product
f × u × v within a square in the right hand top corner for each cell. For example for u = – 1 and v = 2, the
cell frequency f is 10. Therefore, the product of f, u and v is (– 1) × (2) × 10 = – 20 which is written within a
square on the right hand top of cell. Similarly, for u = 2 and v = 1, the product fuv = 0 × 2 × 1 = 0, and so on
for all the remaining cell frequencies.
7. Add together all the figures in the top corner squares as obtained in step 6 to get the last column fuv
for each of the X and Y series. Finally, find the total of the last column to get ∑fuv.
8. Multiply the values of fu and fv by the corresponding values of u and v respectively to get the
columns for fu2 and fv2. Add these values to obtain ∑fu2 and ∑fv2.
The above calculations are shown in the table given on next page 8.28.
N∑fuv – (∑fu) (∑fv)
ruv =
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
[N∑fu2 – (∑fu) 2 ] × [N∑fv2 – (∑fv) 2 ]
=
=
100 × (±48) ± 0 × 100
2
(100 × 120 ± 0) × [100 × 200 ± (100) ]
±4800
12000 × 10000
Hence,
=
±48
120 × 100
=
=
±48
12000
± 4800
[From page 8·28]
12000 × (20000 ± 10000)
– 48
= 109·54
= – 0·4382
rxy = ruv = – 0·4381
[By Property II of r]
Example. 8·22. Calculate Karl Pearson’s coefficient of correlation from the data given below :
Marks
Age in Years
18
19
20
21
22
20—25
3
2
—
—
—
15—20
—
5
4
—
—
10—15
—
—
7
10
—
5—10
—
—
—
3
2
0—5
—
—
—
3
1
Solution. If we denote the age in years by the variable X and the mid-point of the class intervals of
marks by the variable Y and take
u = X – 20; and v =
Y – 12·5
,
5
then the bivariate correlation table is as given on page 8.29.
CALCULATIONS FOR CORRELATION COEFFICIENT
X→
200—300
Mid
pt. x
Y↓
Mid pt. y
U
→
V↓
300—400
400—500
500—600
600—700
250
350
450
550
650
–2
–1
0
1
2
0
10—15
12·5
–1
15—20
17·5
0
20—25
22·5
1
7
25—30
27·5
2
3
0
0
–3
3
0
0
4
– 14
0
9
–6
6
– 12
– 20
10
0
0
12
fv2
fuv
10
– 10
10
–17
20
0
0
0
30
30
30
– 15
40
80
160
– 16
0
3
5
0
5
0
19
fv
– 14
7
4
f
16
0
8
f
10
20
40
20
10
N = 100
fu
– 20
– 20
0
20
20
∑fu = 0
fu2
40
20
0
20
40
∑fu2 = 120
fuv
– 26
– 26
0
18
– 14
∑fuv = – 48
∑fv
= 100
∑fv2
= 200
∑fuv
= – 48
CALCULATIONS FOR CORRELATION COEFFICIENT
Age in
years →
X
Marks
Y↓
u
18
19
20
21
22
–2
–1
0
1
2
v
–12
22·5
2
3
12·5
7·5
2·5
0
0
0
–5
0
0
0
0
0
0
–3
–4
2
0
17·5
–4
1
5
4
0
0
0
0
0
0
0
0
0
7
19
–1
3
–2
Total f
2
–6
Total
f
fv
fv2
fuv
5
10
20
–16
9
9
9
–5
17
0
0
0
5
–5
5
–7
4
–8
16
– 10
∑fv = 6
∑fv2 = 50
∑fuv
= – 38
–4
3
1
3
7
11
16
3
N = 40
fu
–6
–7
0
16
6
∑fu = 9
fu2
12
7
0
16
12
∑fu2 = 47
fuv
– 12
–9
0
–9
–8
∑fuv = – 38
8·30
BUSINESS STATISTICS
ruv =
=
N∑fuv – (∑fu)(∑fv)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
–
×
–
[N∑fu2
(∑fu) 2 ]
– 1520 – 54
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
(1880 – 81) (2000 – 36)
[N∑fv2
=
(∑fv) 2 ]
– 1574
⎯
√⎯⎯⎯⎯⎯⎯⎯⎯⎯
1799 × 1964
=
=
40 × (±38) ± 9 × 6
[From page 8·29]
[ 40 × 47 ± (9)2 ] × [40 × 50 ± (6)2 ]
– 1574
⎯⎯⎯⎯⎯⎯
√
3533236
– 1574
= 1879·69
= 0·8373
But since correlation coefficient is independent of change of origin and scale, [c.f. Property II of r], we
get
rxy = ruv = – 0·8373.
EXERCISE 8·3
1. Write a brief note on the correlation table.
The following are the marks obtained by 24 students in a class test of Statistics and Mathematics :
Roll No. of Students
:
1
2
3
4
5
6
7
8
9
10
11
12
Marks in Statistics
:
15
0
1
3
16
2
18
5
4
17
6
19
Marks in Mathematics :
13
1
2
7
8
9
12
9
17
16
6
18
Roll No. of Students
:
13
14
15
16
17
18
19
20
21
22
23
24
Marks in Statistics
:
14
9
8
13
10
13
11
11
12
18
9
7
Marks in Mathematics :
11
3
5
4
10
11
14
7
18
15
15
3
Prepare a correlation table taking the magnitude of each class interval as four marks and the first class interval as
“equal to 0 and less than 4”. Calculate Karl Pearson’s coefficient of correlation between the marks in Statistics and
marks in Mathematics from the correlation table and comment on it.
Ans. r = 0·5717.
2. What is a bivariate table ? Write the formula you use for calculating coefficient of correlation from such a table,
explaining the symbols used. What does a negative value of the coefficient of correlation indicate ?
3. Calculate the coefficient of correlation between the ages of husbands and wives and its probable error from the
following table :
Ages of husbands (years)
Ages of wives (years)
20—30
30—40
40—50
50—60
60—70
Total
15—25
25—35
35—45
45—55
55—65
5
—
—
—
—
9
10
1
—
—
3
25
12
4
—
—
2
2
16
4
—
—
—
5
2
17
37
15
25
6
Total
5
20
44
24
7
100
Ans. r = 0·7953; P.E. (r) = 0·0248.
Y
4. (a) Given the data in the adjoining table, calculate
the correlation coefficient, r, between X and Y.
30—50
50—70
70—90
X
0—5
5—10
10—15
10
3
4
6
5
7
2
4
9
[Delhi Univ. B.A. (Econ. Hons.), 1991]
Ans. r = 0·375.
(b) Given the following table, obtain r xy.
x
0—5
5—10
10—15
Total
y
30—50
10
3
4
17
50—70
6
—
7
—
70—90
2
4
—
15
Total
18
12
20
50
[Delhi Univ. B.A. (Econ. Hons.), 2004]
Hint. First, obtain the missing values from given totals
Ans. r = 0·375.
8·31
CORRELATION ANALYSIS
x
5
10
15
20
y
5. Calculate the product moment coefficient of
correlation for the adjoining bivariate distribution.
11
2
4
5
4
17
5
3
6
2
Ans. r = – 0·0856.
23
3
1
2
3
6. The following table gives the distribution of income and expenditure of 100 families. Find the coefficient of
correlation and its probable error. State whether the correlation coefficient is significant or not.
Income (in ’00 Rs.)
Percentage Expenditure on Food (X)
Y
10—20
20—30
30—40
40—50
50—60
350—450
—
—
—
—
5
450—550
—
—
1
10
9
550—650
—
4
12
25
3
650—750
4
16
2
2
—
750—850
2
5
—
—
—
Ans. – 0·8248.
7. A sample of 100 firms was taken and these were classified according to the sales executed by them and profits
earned consequently. The results are shown in the table given below. Determine the correlation coefficient between
sales and profits and also its probable error.
Profits
Sales (in lakhs of Rs.)
(in ’000 Rs.)
7—8
8—9
9—10
10—11
11—12
12—13
Total
50—70
5
3
—
—
—
—
8
70—90
3
8
5
4
—
—
20
90—110
1
—
7
11
2
2
23
110—130
—
4
5
15
6
—
30
130—150
—
—
2
7
4
6
19
Total
9
15
19
37
12
8
100
Ans. r = 0·6627; P.E.(r) = 0·0378.
8·7. RANK CORRELATION METHOD
Sometimes we come across statistical series in which the variables under consideration are not capable
of quantitative measurement but can be arranged in serial order. This happens when we are dealing with
qualitative characteristics (attributes) such as honesty, beauty, character, morality, etc., which cannot be
measured quantitatively but can be arranged serially. In such situations Karl Pearson’s coefficient of
correlation cannot be used as such. Charles Edward Spearman, a British psychologist, developed a formula
in 1904 which consists in obtaining the correlation coefficient between the ranks of n individuals in the two
attributes under study.
Suppose we want to find if two characteristics A, say, intelligence and B, say, beauty are related or not.
Both the characteristics are incapable of quantitative measurements but we can arrange a group of n
individuals in order of merit (ranks) w.r.t. proficiency in the two characteristics. Let the random variables X
and Y denote the ranks of the individuals in the characteristics A and B respectively. If we assume that there
is no tie, i.e., if no two individuals get the same rank in a characteristic then, obviously, X and Y assume
numerical values ranging from 1 to n.
The Pearsonian correlation coefficient between the ranks X and Y is called the rank correlation
coefficient between the characteristics A and B for that group of individuals.
Spearman’s rank correlation coefficient, usually denoted by ρ (Rho) is given by the formula
6∑d2
ρ=1–
… (8·15)
n(n2 – 1)
where d is the difference between the pair of ranks of the same individual in the two characteristics and n is
the number of pairs.
8·7·1. Limits for ρ. Spearman’s rank correlation coefficient lies between –1 and +1, i.e.,
–1 ≤ ρ ≤ 1
…(8·16)
8·32
BUSINESS STATISTICS
Remark. Since the square of a real quantity is always non-negative, i.e., ≥ 0, ∑d2 being the sum of
non-negative quantities is also non-negative. Further since n is also positive we get from (8·15)
ρ = 1 – [some non-negative quantity]
⇒
ρ ≤ 1,
…(i)
2
2
the sign of equality holds i.e., ρ = 1 if and only if ∑d = 0. Now, ∑d = 0 if and only if each d = 0, i.e., the
ranks of an individual are same in both the characteristics. Following table gives one such possibility.
x
1
2
3
…
n
y
1
2
3
…
n
On the other hand, ρ will be minimum i.e., ρ = –1 if ∑d2 is maximum, i.e., if the deviations d are
maximum which is so if the ranks of the individuals in the two characteristics are in the reverse (opposite)
order as given in the following table.
Individual
1
2
3
…
n–1
n
x
1
2
3
…
n–1
n
y
n
n–1
n–2
…
2
1
Note. From the above table, we observe that : x + y = n + 1. Also x – y = d
8·7·2. Computation of Rank Correlation Coefficient. We shall discuss below the method of
computing the Spearman’s rank correlation coefficient ρ under the following situations :
(i) When actual ranks are given.
(ii) When ranks are not given.
CASE (I) — WHEN ACTUAL RANKS ARE GIVEN
In this situation the following steps are involved :
(i) Compute d, the difference of ranks.
(ii) Compute d2 .
(iii) Obtain the sum ∑d2 .
(iv) Use formula (8·15) to get the value of ρ.
Example 8·23. The ranks of the same 15 students in two subjects A and B are given below ; the two
numbers within the brackets denoting the ranks of the same student in A and B respectively. (1, 10), (2, 7),
(3, 2), (4, 6), (5, 4), (6, 8), (7, 3), (8, 1), (9, 11), (10, 15), (11, 9), (12, 5), (13, 14), (14, 12), (15, 13).
Use Spearman’s formula to find the rank correlation coefficient.
[Sukhadia Univ. MBA, 1998]
Solution.
CALCULATION OF SPEARMAN’S
Spearman’s rank correlation coefficient ρ is given
by :
CORRELATION COEFFICIENT
Rank in B
d=x–y
d2
Rank in A
(y)
(x)
——————————————————
——————————————————
———————————————————
—————————————————
ρ =1–
6∑ d2
n(n2 – 1)
6 × 272
= 1 – 15(225
– 1)
6 × 272
= 1 – 15
× 224
18
= 1 – 17
35 = 35
= 0·51
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
10
7
2
6
4
8
3
1
11
15
9
5
14
12
13
–9
–5
1
–2
1
–2
4
7
–2
–5
2
7
–1
2
2
∑d=0
81
25
1
4
1
4
16
49
4
25
4
49
1
4
4
∑ d 2 = 272
8·33
CORRELATION ANALYSIS
Example 8·24. Ten competitors in a beauty contest are ranked by three judges in the following order :
1st Judge
2nd Judge
3rd Judge
:
:
:
1
3
6
6
5
4
5
8
9
10
4
8
3
7
1
2
10
2
4
2
3
9
1
10
7
6
5
8
9
7
Use the rank correlation coefficient to determine which pair of judges has the nearest approach to
common tastes in beauty.
Solution. Let R1 , R2 and R 3 denote the ranks given by the first, second and third judges respectively
and let ρij be the rank correlation coefficient between the ranks given by ith and jth judges, i ≠ j = 1, 2, 3.
Let dij = Ri – Rj, be the difference of ranks of an individual given by the ith and jth judge.
CALCULATION OF RANK CORRELATION COEFFICIENT
R1
R2
R3
1
6
5
10
3
2
4
9
7
8
3
5
8
4
7
10
2
1
6
9
6
4
9
8
1
2
3
10
5
7
d 12 = R1 – R2 d 13 = R1 – R3 d 23 = R2 – R3
–3
1
–1
–4
6
8
–1
–9
1
2
–5
2
–4
2
2
0
1
–1
2
1
–2
1
–3
6
–4
–8
2
8
1
–1
∑ d12 = 0
∑ d13 = 0
∑ d23 = 0
d 122
d 132
d 232
4
1
9
36
16
64
4
64
1
1
25
4
16
4
4
0
1
1
4
1
9
1
1
16
36
64
1
81
1
4
∑ d122 = 200
∑ d132 = 60
∑ d232 = 214
We have n = 10.
Spearman’s rank correlation coefficients are given by :
6∑d122
6 × 200 = – 7
ρ12 = 1 –
= 1 – 10
= – 0·2121
× 99
33
n(n2 – 1)
6∑d132
6 × 60
7
ρ13 = 1 –
= 1 – 10
= 0·6363
× 99 = 11
n(n2 – 1)
6∑d232
6 × 214 = – 49 = – 0·2970
ρ23 = 1 –
= 1 – 10
× 99
165
n(n2 – 1)
Since ρ 13 is maximum, the pair of first and third judges has the nearest approach to common tastes in
beauty.
Remark. Since ρ12 and ρ23 are negative, the pair of judges (1, 2) and (2, 3) have opposite (divergent)
tastes for beauty.
CASE (II)—WHEN RANKS ARE NOT GIVEN
Spearman’s rank correlation formula (8·15) can also be used even if we are dealing with variables
which are measured quantitatively, i.e., when the actual data but not the ranks relating to two variables are
given. In such a case we shall have to convert the data into ranks. The highest (smallest) observation is
given the rank 1. The next highest (next lowest) observation is given rank 2 and so on. It is immaterial in
which way (descending or ascending) the ranks are assigned. However, the same approach should be
followed for all the variables under consideration.
Example 8·25. Calculate Spearman’s rank correlation coefficient between advertisement cost and
sales from the following data :
Advertisement cost (’000 Rs.) :
Sales (lakhs Rs.)
:
39
47
65
53
62
58
90
86
82
62
75
68
25
60
98
91
36
51
Solution. Let X denote the advertisement cost (’000 Rs.) and Y denote the sales (lakhs Rs.).
78
84
8·34
BUSINESS STATISTICS
CALCULATION OF RANK CORRELATION COEFFICIENT
Y
Rank of X (x)
Rank of Y (y)
d=x–y
–2
10
8
47
–2
8
6
53
0
7
7
58
0
2
2
86
–2
5
3
62
1
4
5
68
4
6
10
60
0
1
1
91
0
9
9
51
1
3
4
84
∑d=0
X
39
65
62
90
82
75
25
98
36
78
d2
4
4
0
0
4
1
16
0
0
1
∑ d 2 = 30
Here n = 10
6∑d 2
6 × 30
2
9
= 1 – 10
× 99 = 1 – 11 = 11 = 0·82.
n (n2 – 1)
Example 8·26. A test in Statistics was taken by 7 students. The teacher ranked his pupils according to
their academic achievement. The order of achievement from high to low, together with family income for
each pupil, is given as follows :
Rai (Rs. 8,700), Bhatnagar (Rs. 4,200), Tuli (Rs. 5,700), Desai (Rs. 8,200), Gupta (Rs. 20,000),
Chaudhri (Rs. 18,000) and Singh (Rs. 17,500).
Compute the Spearman’s rank correlation between academic achievement and family income.
[Delhi Univ. B.Com. (Pass), 1999]
Solution. Let us define the following variables.
∴
ρ =1–
X : Academic achievement ; Y : Family income (Rupees)
CALCULATIONS FOR RANK CORRELATION COEFFICIENT
Student
Rai
Bhatnagar
Tuli
Desai
Gupta
Chaudhri
Singh
Rank X (x)
Income (Rs.) Y
1
2
3
4
5
6
7
8,700
4,200
5,700
8,200
20,000
18,000
17,500
Rank Y (y)
4
7
6
5
1
2
3
d=x–y
d2
–3
–5
–3
–1
4
4
4
∑d=0
9
25
9
1
16
16
16
∑d 2 = 92
Spearman’s rank correlation coefficient between academic achievement (X) and family income (Y) is
given by :
6 ∑d2
92
9
ρxy = 1 –
= 1 – 76 ×× 48
= 1 – 23
= – 14
= – 0·6429.
14
n(n2 – 1)
REPEATED RANKS
In case of attributes if there is a tie i.e., if any two or more individuals are placed together in any
classification w.r.t. an attribute or if in case of variable data there is more than one item with the same value
in either or both the series, then Spearman’s formula (8·15) for calculating the rank correlation coefficient
breaks down, since in this case the variables X [the ranks of individuals in characteristic A (1st series)] and
Y [the ranks of individuals in characteristic B (2nd series)] do not take the values from 1 to n and
–
–
consequently –x ≠ y, while in proving (8·15) we had assumed that –x = y .
8·35
CORRELATION ANALYSIS
In this case, common ranks are assigned to the repeated items. These common ranks are the arithmetic
mean of the ranks which these items would have got if they were different from each other and the next
item will get the rank next to the rank used in computing the common rank. For example, suppose an item
is repeated at rank 4. Then the common rank to be assigned to each item is (4 + 5)/2, i.e., 4·5 which is the
average of 4 and 5, the ranks which these observations would have assumed if they were different. The next
item will be assigned the rank 6. If an item is repeated thrice at rank 7, then the common rank to be
assigned to each value will be (7 + 8 + 9)/3, i.e., 8 which is the arithmetic mean of 7, 8 and 9, viz., the ranks
these observations would have got if they were different from each other. The next rank to be assigned will
be 10.
If only a small proportion of the ranks are tied, this technique may be applied together with formula
(8·15). If a large proportion of ranks are tied, it is advisable to apply an adjustment or a correction factor
(C.F. ) to (8·15) as explained below.
“In the formula (8·15) add the factor m (m2 – 1)/12 to ∑d2 , where m is the number of times an item is
repeated. This correction factor is to be added for each repeated value in both the series.”
Example 8·27. A psychologist wanted to compare two methods A and B of teaching. He selected a
random sample of 22 students. He grouped them into 11 pairs so that the students in a pair have
approximately equal scrores on an intelligence test. In each pair one student was taught by method A and
the other by method B and examined after the course. The marks obtained by them are tabulated below :
Pair :
1
2
3
4
5
6
7
8
9
10
11
A
:
24
29
19
14
30
19
27
30
20
28
11
B
:
37
35
16
26
23
27
19
20
16
11
21
Find the rank correlation coefficient.
Solution. Let variable X denote the scores of students taught by method A and Y denote the scores of
students taught by method B.
CALCULATIONS FOR RANK CORRELATION COEFFICIENT
Rank of X Rank of Y d = x – y
In the X-series, we see that the value
X
Y
d2
(x)
(y)
30 occurs twice. The common rank
——————————————————————————————————————
————————————————
—————————————————
—————————————————————
assigned to each of these values is 1·5, the
arithmetic mean of 1 and 2, the ranks
which these observations would have
taken if they were different. The next
value 29 gets the next rank, viz., 3. Again,
the value 19 occurs twice. The common
rank assigned to it is 8·5, the arithmetic
mean of 8 and 9 and the next value, viz.,
14 gets the rank 10. Similarly, in the
y-series the value 16 occurs twice and the
common rank assigned to each is 9·5, the
arithmetic mean of 9 and 10. The next
value viz., 11 gets the rank 11.
24
29
19
14
30
19
27
30
20
28
11
37
35
16
26
23
27
19
20
16
11
21
6
3
8·5
10
1·5
8·5
5
1·5
7
4
11
1
2
9·5
4
5
3
8
7
9·5
11
6
5
1
–1
6
–3·5
5·5
–3
–5·5
–2·5
–7
5
25
1
1
36
12·25
30·25
9·00
30·25
6·25
49·00
25·00
————————————
————————————
————————————————
————————————————
————————————————
——————————————————————
∑d = 0
∑d2 = 225·00
Hence, we see that in the X-series the items 19 and 30 are repeated, each occurring twice and in the
Y-series the item 16 is repeated. Thus in each of the three cases m = 2. Hence on applying the correction
m(m2 – 1)
factor
for each repeated item, we get
12
6
ρ =1–
[
∑d2 +
2(4 – 1) 2(4 – 1) 2(4 – 1)
12 + 12 + 12
11(121 – 1)
]
= 1 – 611× ×226·5
120 = 1 – 1·0225 = – 0·0225.
(·. · n = 11)
8·36
BUSINESS STATISTICS
Example 8·28. From the following data calculate the rank correlation coefficient after making
adjustment for tied ranks.
x:
48
33
40
9
16
16
65
24
16
57
y:
13
13
24
6
15
4
20
9
6
19
[I.C.W.A. (Intermediate), June 2002]
Solution. Rx = Rank of x-value ; Ry = Rank of y-value
x
Explanation of Ranks
y
Rx
Ry
d2
d = Rx – Ry
—————————————
—————————————————————————————
——————————————
—————————————————
—————————————————
48
33
40
9
16
16
65
24
16
57
In the x-series, we see that the value 16
is repeated three times. The common rank
assigned to each of these values is 8, the
arithmetic mean of 7, 8 and 9, the ranks
which these observations would have taken
if they were different. The next value 9 gets
the next rank viz., 10
13
13
24
6
15
4
20
9
6
19
3
5
4
10
8
8
1
6
8
2
5·5
5·5
1
8·5
4
10
2
7
8·5
3
–2·5
– 0·5
3
1·5
4
–2
–1
–1
– 0·5
–1
6·25
0·25
9
2·25
16
4
1
1
0·25
1
Similarly, in the y-series the observation
13 occurs twice. The common rank assigned
to each is 5·5, the arithmetic mean of 5 and
∑d = 0
∑d 2 = 41
6, and the next value 9 gets the next rank 7. Total
Again the value 6 is repeated twice and each is given the common rank (8 + 9)/2 i.e., 8·5 and the next value
4 gets the rank 10.
Correction Factor (C.F.)
Repeated value No. of times it
C.F.
2– 1)/12
occurs
(m)
m(m
Spearman’s rank correlation coefficient for
repeated ranks gives :
x–series : 16
3
3 (9 – 1)/12 = 2
————————————
————————————
————————————————
—————————————————————————————
——————————————————
———————————————————————
——————————————————————————————————————————————
ρ =1–
6
– 1)
n(n2
[ ∑ d + ∑ m(m12– 1) ]
2
2
[
6
= 1 – 10(100
– 1) 41 + 3
4
11
= 1 – 15
= 15
= 0·7333.
]=1–
6 × 44
10 × 99
y–series : 13
6
2
2
2(4 – 1)/12 = 0·5
2(4 – 1)/12 = 0·5
—————————————————————————
—————————————————————————————————————————————
——
3
Total
Example 8.29. Find the Rank correlation Coefficient from the following marks awarded by the
examiners in Statistics.
Roll No.
By Examiner A
By Examiner B
By Examiner C
1
2
3
4
5
6
7
8
9
10
11
24
29
19
14
30
19
27
30
20
28
11
37
35
16
26
23
27
19
20
16
11
21
30
28
20
25
25
30
20
24
22
29
15
[Delhi Univ. B.Com. (Hons.), 2005]
8·37
CORRELATION ANALYSIS
Solution· First of all we shall convert the markets awarded by different examiners to the candidates
into ranks, as given in the following table.
Rok
Nos·
1
2
3
4
5
6
7
8
9
10
11
Marks
Rank
Marks
awarded (R A) awarded
by
by
Examiner
Examiner
B
A
24
6
37
29
3
35
19
8·5
16
14
10
26
30
1·5
23
19
8·5
27
27
5
19
30
1·5
20
20
7
16
28
4
11
11
11
21
We are given n = 11.
Rank
(R B)
Marks
awarded
by
Examiner
C
Rank
(R C)
1
2
9·5
4
5
3
8
7
9·5
11
6
30
28
20
25
25
30
20
24
22
29
15
1·5
4
9·5
5·5
5·5
1·5
9·5
7
8
3
11
[
n
[
6 225 +
=1–
=1–
5
1
–1
6
– 3·5
5·5
–3
– 5·5
– 2·5
–7
5
4·5
–1
–1
4·5
–4
7
– 4·5
– 5·5
–1
1
0
– 0·5
–2
0
– 1·5
– 0·5
1·5
– 1·5
0
1·5
8
–5
d 2AB
d 2AC
d 2BC
25
1
1
36
12·25
30·25
9
30·25
6·25
49
25
20·25
1
1
20·25
16
49
20·25
30·25
1
1
0
0·25
4
0
2·25
0·25
2·25
2·25
0
2·25
64
25
∑d 2 AB =
225
∑d 2 AC =
160
∑d 2 BC =
102·5
m (m2 – 1)
for each repeated rank in A and B series
12
6 ∑dAB 2 +
ρAB = 1 –
d AB = d AC =
d BC =
RA – RB RA – RC RB – RC
(n2
]
– 1)
2 (22 – 1) 2 (22 – 1) 2 (22 – 1)
+
+
12
12
12
]
11 (121 – 1)
6 × 226·5
226·5
26·5
=1–
=–
= – 0·0295
11 × 120
220
220
Similarly
ρAC = 1 –
6 [∑ dAC2 + correction for Repeated Ranks]
n (n2 – 1)
[
6 160 + 2
=1–
=1–
} {
2 (22 – 1)
+3
12
2 (22 – 1)
12
11 × (121 – 1)
}]
6 (160 + 2·5)
162·5 57·5
=1–
=
= 0·2614
1 × 220
220
220
[
6 102·5 +
ρBC = 1 –
{
2 (22 – 1)
+3
12
{
2 (22 – 1)
12
}]
11 (121 – 1)
6 × 104·5
104·5 115·5
=1–
=
= 0·5250
11 × 120
220
220
Example 8·30. The value of Spearman’s rank correlation coefficient for certain pairs of number of
observations, was found to be 2/3. The sum of squares of the differences between corresponding ranks was
55. Find the number of pairs.
=
8·38
BUSINESS STATISTICS
Solution. We are given ρ = 2/3 and ∑ d2 = 55. If n is the number of pairs of observations, we have :
6∑d2
n(n2 – 1)
2 1
–3=3
ρ=1–
330
=
n(n2 – 1)
⇒
1
6 × 55
n(n2 – 1)
⇒
2
3
⇒
n(n2 – 1) = 3 × 330 = 990
=1
–
⇒
n3 – n – 990 = 0
…(*)
By hit and trial we find that on putting n = 10 in (*),
L.H.S. = 103 – 10 – 990 = 1000 – 1000 = 0 = R.H.S.
Hence, by Remainder Theorem, (n – 10) is a factor of n 3 – n – 990. On dividing n3 – n – 990 by
(n – 10), we obtain the other factor of n3 – n – 990 as n2 + 10n + 99.
∴
(*)
⇒
(n – 10) (n2 + 10n + 99) = 0
⇒
n – 10 = 0
or
n2 + 10n + 99 = 0
⇒
n = 10
100 – 396
n = –10 ± ⎯√⎯⎯⎯⎯⎯⎯
(Imaginary values)
2
or
Hence, n = 10 is the only permissible value. Hence, the number of pairs is 10.
Example 8·31. The coefficient of rank correlation between Micro-Economics and Statistics marks of
10 students was found to be 0·5. It was later discovered that the difference in ranks in two subjects obtained
by one of the students was wrongly taken as 3 instead of 7. Find the correct value of coefficient of rank
correlation.
[Delhi Univ. B.A. (Econ. Hons.), 2006]
Solution. We are given n = 10, ρ = 0·5. Using (8·15), we get
2
6∑d2
6∑d2
0·5 = 1 –
= 1 – 106∑d
⇒
⇒
∑d2 = 6990
× 99
990 = 1 – 0·5 = 0·5
× 2 = 82·5
n(n2 – 1)
Since one difference was wrongly taken as 3 instead of 7, the correct value of ∑d2 is given by :
Corrected ∑d2 = 82·5 – 3 2 + 7 2 = 82·5 – 9 + 49 = 122·5
∴
Corrected ρ = 1 –
6 × 122·5
10 × 99
= 1–
49
66
= 1 – 0·7424 = 0·2576
8·7·3. Remarks on Spearman’s Rank Correlation Coefficient
1. We always have ∑d = 0, which provides a check for numerical calculations.
2. Since Spearman’s rank correlation coefficient ρ is nothing but Pearsonian correlation coefficient
between the ranks, it can be interpreted in the same way as the Karl Pearson’s correlation coefficient.
3. Karl Pearson’s correlation coefficient assumes that the parent population from which sample
observations are drawn is normal. If this assumption is violated then we need a measure which is
distribution-free (or non-parametric). A distribution-free measure is one which does not make any
assumptions about the parameters of the population. Spearman’s ρ is such a measure (i.e., distributionfree), since no strict assumptions are made about the form of the population from which sample
observations are drawn.
4. Spearman’s formula is easy to understand and apply as compared with Karl Pearson’s formula. The
values obtained by the two formulae, viz., Pearsonian r and Spearman’s ρ are generally different. The
difference arises due to the fact that when ranking is used instead of full set of observations, there is always
some loss of information. Unless many ties exist, the coefficient of rank correlation should be only slightly
lower than the Pearsonian coefficient.
5. Spearman’s formula is the only formula to be used for finding correlation coefficient if we are
dealing with qualitative characteristics which cannot be measured quantitatively but can be arranged
serially. It can also be used where actual data are given. In case of extreme observations, Spearman’s
formula is preferred to Pearson’s formula.
6. Spearman’s formula has its limitations also. It is not practicable in the case of bivariate frequency
distribution (Correlation Table). For n > 30, this formula should not be used unless the ranks are given,
since in the contrary case the calculations are quite time consuming.
8·39
CORRELATION ANALYSIS
EXERCISE 8·4
1. (a) What is Spearman’s rank correlation coefficient ? Discuss its usefulness.
(b) Explain the difference between Karl Pearson’s (product moment) correlation coefficient and rank correlation
coefficient.
2. (a ) What are the advantages of Spearman’s rank correlation coefficient over Karl Pearson’s correlation
coefficient ? Explain the method of calculating Spearman’s correlation coefficient.
(b) Define rank correlation coefficient. When is it preferred to Karl Pearson’s coefficient of correlation ?
(c) Distinguish between Karl Pearson’s coefficient of correlation and Spearman’s rank correlation coefficient.
Explain with the help of an example when Spearman rank correlation coefficient results to + 1, – 1 and between – 1 to
+ 1.
[Delhi Univ. B.Com. (Hons.), 2009]
3. Define Rank Correlation. Write down Spearman’s formula for rank correlation coefficient ρ. What are the limits
of ρ ? Interpret the case when ρ assumes the minimum value.
4. Rankings of 10 trainees at the beginning (x) and at the end (y) of a certain course are given below :
Trainees
:
A
B
C
D
E
F
G
H
I
x
:
1
6
3
9
5
2
7
10
8
y
:
6
8
3
7
2
1
5
9
4
Calculate Spearman’s rank correlation coefficient.
J
4
10
[I.C.W.A. (Intermediate), June 1995]
Ans. ρ = 0·394.
5. The ranks of same 16 students in Mathematics and Physics are as follows. Two numbers within brackets denote
the ranks of the students in Mathematics and Physics.
(1, 1) (2, 10) (3, 3) (4, 4) (5, 5) (6, 7) (7, 2) (8, 6) (9, 8) (10, 11) (11, 15) (12, 9) (13, 14) (14, 12) (15, 16) (16, 13).
Calculate the rank correlation coefficient for proficiencies of this group in Mathematics and Physics.
Ans. ρ = 0·8.
6. Two judges in a beauty competition rank the 12 entries as follows :
X
:
1
2
3
4
5
6
7
Y
:
12
9
6
10
3
5
4
8
7
9
8
10
2
11
11
12
1
What degree of agreement is there between the two judges.
Ans. ρ = – 0·454
7. Ten competitors in a beauty contest are ranked by three judges in the following order :
1st Judge
:
1
5
4
8
9
6
10
7
2nd Judge
:
4
8
7
6
5
9
10
3
3rd Judge
:
6
7
8
1
5
10
9
2
3
2
3
2
1
4
Use the rank correlation coefficient to discuss which pair of judges has the nearest approach to beauty.
Ans. ρ12 = 0·5515,
ρ13 = 0·0545,
ρ23 = 0·7333.
The pair of 2nd and 3rd judges has the nearest approach to common tastes in beauty.
8. For the following data, calculate the Coefficient of Rank Correlation.
x
:
80
91
99
71
61
y
:
123
135
154
110
105
Ans. ρ = 0·9524.
81
134
70
121
59
106
[C.A. (Foundation), May 2001]
9. The following are the marks obtained by a group of students in two papers. Calculate the rank coefficient of
correlation.
Economics
:
78
36
98
25
75
82
92
62
65
39
Statistics
:
84
51
91
69
68
62
86
58
35
49
Ans. ρ = 0·6121.
10. Calculate Spearman’s coefficient of rank correlation for the following data of scores in psychological tests (x)
and arithmetical ability (y) of 10 children.
8·40
BUSINESS STATISTICS
Child :
x :
y :
A
105
101
B
104
103
C
102
100
D
101
98
E
100
95
F
99
96
G
98
104
H
96
92
I
93
97
J
92
94
Ans. ρ = 0·6.
11. How do you modify Spearman’s rank correlation formula for tied ranks ? Compute the Coefficient of Rank
Correlation between X and Y from the data given below :
X :
8
10
7
15
3
20
21
5
10
14
8
16
22
19
6
Y :
3
12
8
13
20
9
14
11
4
16
15
10
18
23
25
Ans. ρ = 1 –
6 × (539 + 0·5 + 0·5)
15 × 224
= 0·0357.
[Delhi Univ. B.Com. (Pass), 1998]
12. Given the following aptitude and I.Q. scores for a group of students. Find the coefficient of rank correlation.
Aptitude Score
:
57
58
59
59
60
61
60
64
I.Q. Score
:
97
108
95
106
120
126
113
110
[Delhi Univ. B.A. (Eon. Hons.), 2009]
6 × (24 + 0.5 = 0.5)
150
Ans. ρ = 1 –
=1–
= 0.7024
8 (64 – 1)
504
13. The following data relate to the marks obtained by 10 students of a class in Statistics and Costing :
Marks in Statistics
:
30
38
28
27
28
23
30
33
28
35
Marks in Costing
:
29
27
22
29
20
29
18
21
27
22
Obtain the rank correlation coefficient.
[Delhi Univ. B.Com. (Hons.), 2001]
Ans. ρ = – 0·3515.
14. Find the coefficient of rank correlation between the marks obtained in Mathematics (x) and those in Statistics
(y) by 10 students of certain class out of a total of 50 marks in each subject.
Student No. :
1
2
3
4
5
6
7
8
9
10
x :
12
18
32
18
25
24
25
40
38
22
y :
16
15
28
16
24
22
28
36
34
19
Ans. ρ = 0·95.
15. From the following data, calculate the coefficient of rank correlation between x and y.
x
:
32
35
49
60
43
37
43
49
y
:
40
30
70
20
30
50
72
60
10
45
20
25
Ans. ρ = – 0·0758.
16. When is rank correlation coefficient preferred to Karl Pearson’s method ? In a bivariate sample, the sum of
squares of differences between the ranks of observed values of two variables is 231 and the correlation coefficient
between them is – 0·4. Find the number of pairs.
[Delhi Univ. B.Com. (Hons.) (External), 2006]
Ans. n = 10.
Hint. ρ = 1 –
⇒
6 ∑d 2
n (n2 – 1)
n 3 – n – 990 = 0
⇒
n (n 2 – 1) =
⇒
6 ∑ d2 6 × 231
=
= 990
1·4
1–ρ
n = 10 [See Example 8.30]
17. If the rank correlation coefficient is 0·6 and the sum of the squares of differences of ranks is 66, then the
number of pairs is
(i) 8,
(ii) 9, (iii) 10,
(iv) 11.
Ans. (iii).
[I.C.W.A. (Intermediate), June 2001]
18. Coefficient of correlation between debenture prices and share prices is found to be 0·143. If the sum of the
squares of differences in ranks is given to be 48, find the value of n.
Ans. n = 7.
19. The coefficient of rank correlation of the marks obtained by 10 students in biology and chemistry was found to
be 0·8. It was later discovered that the difference in ranks in the two subjects obtained by one of the students was
wrongly taken as 7 instead of 9. Find the correct coefficient of rank correlation.
8·41
CORRELATION ANALYSIS
Ans. Correct value of ρ = 0·6061.
20. The coefficient of rank correlation of the marks obtained by 10 students in Statistics and Accountancy was
found to be 0·2. It was later discovered that the difference in ranks in the two subjects obtained by one of the students
was wrongly taken as 9 instead of 7. Find the correct value of coefficient of rank correlation.
[Delhi Univ. B.Com (Hons.), 1992]
Ans. Correct (ρ) = 0·3939.
21. The rank correlation of a physical fitness contest involving 12 participants was calculated as 0·6. However, it
was later discovered that the difference in ranks of a participant was read as 8 instead of 3. Find the correct value of
rank correlation.
[Delhi Univ. B.A. (Econ. Hons.), 2009]
Ans. 0·7924.
22. Mention the correct answer.
The ranks according to two attributes in a sample are given below :
R1
:
1
2
3
R2
:
5
4
3
The rank correlation between them is :
0,
+1,
–1,
Ans. ρ = –1.
4
2
5
1
None of these.
8·8. METHOD OF CONCURRENT DEVIATIONS
This is very casual method of determining the correlation between two series when we are not very
serious about its precision. This is based on the signs of the deviations (i.e., direction of the change) of the
values of the variable from its preceding value and does not take into account the exact magnitude of the
values of the variables. Thus we put a plus (+) sign, minus (–) sign or equality (=) sign for the deviation if
the value of the variable is greater than, less than or equal to the preceding value respectively. The
deviations in the values of two variables are said to be concurrent if they have the same sign, i.e., either
both deviations are positive or both are negative or both are equal. The formula used for computing
correlation coefficient r by this method is given by
r=±
±
( 2cn– n)
…(8·17)
where c is the number of pairs of concurrent deviations and n is the number of pairs of deviations. In the
formula (8·17) plus/minus sign to be taken inside and outside the square root is of fundamental importance.
2c – n
Since –1 ≤ r ≤ 1, the quantity inside the square root, viz., ±
must be positive, otherwise r will be
n
imaginary which is not possible.
Thus if (2c – n) is positive, we take positive sign in and outside the square root in (8·17) and if (2c – n)
is negative, we take negative sign in and outside the square root in (8·17).
Remarks 1. It should be clearly noted that here n is not the number of pairs of observations but it is the
number of pairs of deviations and as such it is one less than the number of pairs of observations.
2. r computed by formula (8·17) is also known as coefficient of concurrent deviations.
3. Coefficient of concurrent deviations is primarily based on the following principle :
“If the short time fluctuations of the time series are positively correlated or in other words, if their
deviations are concurrent, their curves would move in the same direction and would indicate positive
correlation between them.”
Thus r computed from (8·17) ordinarily indicates the relationship between short time fluctuations only.
Example 8·32. Calculate the coefficient of concurrent deviations from the data given below :
Year
:
1993
1994
1995
1996
1997
1998
1999
2000
2001
Supply
:
160
164
172
182
166
170
178
192
186
Price
:
292
280
260
234
266
254
230
190
200
(
)
8·42
BUSINESS STATISTICS
Solution.
Year
1993
1994
1995
1996
1997
1998
1999
2000
2001
CALCULATION OF COEFFICIENT OF CONCURRENT DEVIATIONS
Supply
Sign of deviation from
Price
Sign of deviation from
Product of
preceding value (x)
preceding value (y)
deviations (xy)
292
160
–
–
280
+
164
–
–
260
+
172
–
–
234
+
182
–
+
266
–
166
–
–
254
+
170
–
–
230
+
178
–
–
190
+
192
–
+
200
–
186
Here we have :
n = Number of pairs of deviations = 9 – 1 = 8
c = 0, since there is no pair of deviations having like signs, i.e., since no product deviations xy is
positive.
Coefficient of concurrent deviations is given by
r =±
±
(2cn– n)
=±
±
( )
0–8
8
= ± √⎯⎯⎯⎯
± (–1)
Since 2c – n = – 8, i.e., (negative), we take negative sign inside and outside the square root to get,
r = – √⎯⎯⎯⎯
– (–1) = –1
Hence, there is perfect negative correlation between the supply and the price.
Example 8·33. Calculate the coefficient of concurrent deviations for the following data :
Supply
:
65
40
35
75
63
80
35
20
80
60
50
Demand :
60
55
50
56
30
70
40
35
80
75
80
[C.A. (Foundation), Nov. 1997]
Solution.
CALCULATIONS FOR COEFFICIENT OF CONCURRENT DEVIATIONS
Supply (X)
Sign of the deviation from
preceding value (x)
65
40
35
75
63
80
35
20
80
60
50
Demand (Y)
Sign of the deviation from
preceding value (y)
Product of deviations
(xy)
–
–
+
–
+
–
–
+
–
+
+
+
+
+
+
+
+
+
+
–
60
55
50
56
30
70
40
35
80
75
80
–
–
+
–
+
–
–
+
–
–
Here we have :
n = Number of pairs of deviations = 11 – 1 = 10
c = Number of pairs of deviations having like signs = 9
The coefficient of concurrent deviations is given by :
r =±
±
(2c – n)
n
=±
±
2 × 9 – 10
10
= ± √⎯⎯⎯
± 0·8
Since 2c – n = 8, is positive, we take positive sign inside and outside the square root so that :
r = + ⎯√⎯0·8 = 0·89.
8·43
CORRELATION ANALYSIS
EXERCISE 8·5
1. (a) Explain the method of concurrent deviations for computing the correlation between two variable series.
(b) Give the points of strength and weakness of finding out the relationship between two variables by the method
of concurrent deviations.
2. Obtain the coefficient of correlation between price of rice and rainfall from the data given below by means of
concurrent deviations.
Year
Price of rice (in Rs.) Annual rainfall in
Year
Price of rice (in Rs.)
Annual rainfall
per quintal
centimetres
per quintal
in centimetres
353
196
1965
315
175
1959
333
190
1966
340
160
1960
390
191
1967
350
158
1961
340
195
1968
350
200
1962
380
196
1969
330
198
1963
340
204
1970
335
195
1964
Ans. r = – 0·3015.
3. Calculate the coefficient of correlation by the method of concurrent deviations from the following data :
Year
:
1991
1992
1993
1994
1995
1996
1997
1998
1999
Supply
:
80
82
86
91
83
85
89
96
93
Price
:
146
140
130
117
133
127
115
95
100
Ans. r = –1.
4. Calculate the coefficient of correlation, using the method of concurrent deviations between supply and demand
of an item for a ten year period as given below :
Year
:
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
Supply
:
125
160
164
174
155
170
165
162
172
175
Demand :
115
125
192
190
165
174
124
127
152
169
[C.A. (Foundation), Nov. 1995]
Ans. r = 0·75.
5. Calculate the coefficient of correlation by the concurrent deviations method.
Supply
:
112
125
126
118
118
121
125
125
131
135
Price
:
106
102
102
104
98
96
97
97
95
90
Ans. r = –0·7.
6. Calculate correlation coefficient by concurrent deviations method :
X
:
150
135
90
140
100
Y
:
60
50
100
80
90
Ans. r = – 0·7071.
7. Calculate coefficient of concurrent deviations from the following data :
x :
100
120
135
135
115
110
110
y :
50
40
60
60
80
55
65
Ans. r = 0.
8. Calculate the coefficient of concurrent deviations from the following data :
No. of pairs of observations = 96
No. of pairs of concurrent deviations = 36
Ans. r = – 0·492.
8·9. COEFFICIENT OF DETERMINATION
Coefficient of correlation between two variable series is a measure of linear relationship between them
and indicates the amount of variation of one variable which is associated with or is accounted for by
another variable. A more useful and readily comprehensible measure for this purpose is the coefficient of
determination which gives the percentage variation in the dependent variable that is accounted for by the
independent variable. In other words, the coefficient of determination gives the ratio of the explained
variance to the total variance. The coefficient of determination is given by the square of the correlation
coefficient, i.e., r2 . Thus,
Explained Variance
Coefficient of determination = r2 =
…(8·18)
Total Variance
8·44
BUSINESS STATISTICS
The coefficient of determination is a much useful and better measure for interpreting the value of r.
According to Tuttle :
“The coefficient of correlation has been grossly over-rated and is used entirely too much. Its square,
the coefficient of determination is a much more useful measure of the linear covariation of two variables.
The reader should develop the habit of squaring every correlation coefficient he finds, cited or stated
before coming to any conclusion about the extent of the linear relationship between the two correlated
variables.”
For example, if the value of r = 0·8, we cannot conclude that 80% of the variation in the relative series
(dependent variable) is due to the variation in the subject series (independent variable). But the coefficient
of determination in this case is r2 = 0·64 which implies that only 64% of the variation in the relative series
has been explained by the subject series and the remaining 36% of the variation is due to other factors.
By the same argument while comparing two correlation coefficients, one of which is 0·4 and the other
is 0·8 it is misleading to conclude that the correlation in the second case is twice as high as correlation in
the first case. The coefficient of determination clearly explains this viewpoint, since in the case r = 0·4, the
coefficient of determination is 0·16 and in the case r = 0·8, the coefficient of determination is 0·64, from
which we conclude that correlation in the second case is four times as high as correlation in the first case.
Remarks 1. The above discussion implies that :
“The closeness of the relationship between two variables as determined by correlation coefficient r is
not proportional.”
2. The following table gives the values of the coefficient of determination (r2 ) for different values of r.
r
r
2
0·1
0·2
0·3
0·4
0·5
0·6
0·7
0·8
0·9
1·0
0·01
0·04
0·09
0·16
0·25
0·36
0·49
0·64
0·81
1·00
It may be seen from the above table that as the value of r decreases, r 2 decreases very rapidly except in
two particular cases r = 0 and r = 1 when we get r2 = r.
3. Coefficient of determination is always non-negative and as such it does not tell us about the
direction of the relationship (whether it is positive or negative) between the two series.
4. Coefficient of Non-determination. The ratio of the unexplained variation to the total variation is
called the coefficient of non-determination. It is usually denoted by K2 and is given by the formula :
Un-explained Variance
Explained Variance
=1–
= 1 – r2
…(8·19)
Total Variance
Total Variance
5. Coefficient of Alienation. The coefficient of alienation is given by the square root of the coefficient
of non-determination, i.e., by K as given below :
K2 =
K =±⎯
√⎯⎯⎯
1 – r2.
…(8·19a)
EXERCISE 8·6
1. (a) If the correlation coefficient is 0·7, then what is the coefficient of determination ? Also interpret its value.
[C.A. (Foundation), Nov. 1997]
(b) What is the coefficient of determination ? How is it useful in interpreting the value of an observed correlation
coefficient r ? Explain with the help of an example.
(c) “A high value of the coefficient of determination is neither necessary nor sufficient to ensure a causal
relationship between X and Y.” Explain.
[Delhi Univ. B.A. (Econ. Hons.), 2005]
2. Explain the terms :
(i) Coefficient of non-determination,
(ii) Coefficient of alienation,
and give their physical interpretation.
3. (a) A correlation coefficient of 0·5 does not mean that 50% of the data are explained. Comment.
[Delhi Univ. B.Com. (Hons.), 1998]
Ans. Statement is true. Only 25% of the variation is explained.
(b) Do you agree with the statement : “r = 0·8 implies that 80% of the data are explained.”
CORRELATION ANALYSIS
8·45
Ans. No. Only 64% of the data are explained.
4. The coefficient of correlation between consumption expenditure (c) and disposable income (y) in a study was
found to be + 0·6. What percentage of variation in c is explained by variation in y ?
Ans. 36% of the variation in c is explained by variation in y.
5. (a) A correlation between two variables has a value r = 0·6 and a correlation between other two variables is 0·3.
Does it mean that the first correlation is twice as strong as the second ?
(b) Correlation between two variables has a value of 0·9 and a correlation between other two variables is 0·3. Can
we infer that the first correlation is thrice as strong as the second ? Give reasons.
[Osmania Univ. B.Com., 1997]
Ans. (a) No. (b) No.
6. Comment on the following :
“The closeness of the relationship between two variables as determined by r, the correlation coefficient between
them, is proportional.”
Ans. Statement is wrong.
7. If correlation coefficient between random variables x and y is +ve, comment on the following statements :
(i) Correlation coefficient between –x and –y is +ve.
(ii) We cannot infer about the sign of correlation between x and –y.
(iii) Interpret the result r 2 = 0·64, where r 2 is co-efficient of determination.
[Delhi Univ. B.A. (Econ. Hons.), 2000]
Ans. (i) True, (ii) False. r(x, –y) = – r(x, y) < 0
(iii) 64% of variation in Y is explained by the linear regression of Y on X. [c.f. Chapter 9.]
8. If the correlation coefficient between X and Y is 0.4 and between U and V is 0.8, does this imply that the extent
of association between U and V is twice as that between X and Y ?
[Delhi Univ. B.A. (Econ. Hons.), 1999]
9. Calculate correlation coefficient and coefficient of determination from the following data.
n = 10, ∑X = 140,
∑Y = 150
∑ (X– 10)2 = 180,
∑(Y – 15)2 = 215,
∑(X – 10) (Y – 15) = 60.
[Delhi Univ. B.Com. (Hons.), (External), 2007]
Ans. r = 0·915; Coefficient of determination = r2 = 0·8372
Hint.
U = X – 10,
V = Y – 15,
n = 10
∑U = ∑X – 10 × 10 = 140 – 100 = 40 ;
∑V = ∑Y – 15 × 10 = 150 – 150 = 0
2
2
∑U = 180; ∑V = 215 ;
∑UV = 60·
n ∑UV – (∑U) (∑V)
r XY = rUV =
= 0·915
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
[n ∑U2 – (∑U)2] [n ∑V2 – (∑V)2]
8.10. LAG AND LEAD CORRELATION
When there exists a cause and effect relationship between two time series, it is usually observed that
there is a time lag between the changes in the values of the independent variable (also called the subject
series) and the dependent variable (also called the relative series). Such phenomenon is usually observed in
most of the economic and business time series. For example, the monthly advertisement expenditure of a
firm on its product and the sales of the product have a fairly good degree of positive correlation. However,
the effect of the advertisement expenditure will be felt on the increased sales of the product only after a
certain period of time, which may be 3 or 4 months or even more. This tendency on the part of the effect
(change in the dependent variable or relative) to occur sometimes after the occurrence of the cause (change
in independent variable or subject) is known as ‘lag’.
If it is known that ‘lag’ exists between two time series, it is imperative to make adjustment for it,
before computing the correlation coefficient between the two series otherwise fallacious conclusions will be
drawn.
In order to make allowance for the ‘lag’, it is necessary to determine the ‘time-lag’ i.e., to estimate the
time period which lapses before the change in the dependent variable is affected after a change in the
independent variable. The ‘period of lag’ can be estimated by plotting the two series on a graph paper and
noting the time distance between the peaks/troughs of two curves. If the peak (trough) in dependent
8·46
BUSINESS STATISTICS
variable (sales) comes after k-months of the peak (trough) in the independent variable (advertisement
expenditure), then there is k-month time-lag between the two variables. Here we say that ‘advertisement
expenditure curve’ will lead by k-months and the ‘sales curve’ would lag by k-months.
Illustration. Let us consider the following hypothetical time series values of monthly advertisement
expenditure (in ’000 Rs.) and sales (in ’000 Rs.) of the product of a firm.
Month
Advertisement
expenditure
(’000 Rs.) (x)
Jan.
Feb.
March
April
May
June
Sales
(’000 Rs.)
(y)
x1
x2
x3
x4
x5
x6
y1
y2
y3
y4
y5
y6
Month
July
Aug.
Sept.
Oct.
Nov.
Dec.
x
y
x7
x8
x9
x10
x11
x12
y7
y8
y9
y10
y11
y12
Suppose that ‘advertisement expenditure’ is known to have its effect on the ‘sales’ after 3 months.
After making allowance for this ‘time-lag’ of 3-months, the correct value of correlation coefficient is
obtained on computing Karl Pearson’s correlation coefficient between the following two-series values.
x
y
x1
y4
x2
y5
x3
y6
x4
y7
x5
y8
x6
y9
x7
y10
x8
y11
Note : The study of ‘Lag-correlation’ is specially useful in the study of economic and business time series.
x9
y12
9
Linear Regression
Analysis
9·1. INTRODUCTION
The literal or dictionary meaning of the word ‘Regression’ is ‘stepping back or returning to the
average value’. The term was first used by British biometrician Sir Francis Galton in the later part of the
19th century in connection with some studies he made on estimating the extent to which the stature of the
sons of tall parents reverts or regresses back to the mean stature of the population. He studied the
relationship between the heights of about one thousand fathers and sons and published the results in a paper
‘Regression towards Mediocrity in Hereditary Stature’. The interesting features of his study were :
(i) The tall fathers have tall sons and short fathers have short sons.
(ii) The average height of the sons of group of tall fathers is less than that of the fathers and the average
height of the sons of a group of short fathers is more than that of the fathers.
In other words, Galton’s studies revealed that the off springs of abnormally tall or short parents tend to
revert or step back to the average height of the population, a phenomenon which Galton described as
Regression to Mediocrity.
He concluded that if the average height of a certain group of fathers is ‘a’ cms. above (below) the
general average height then average height of their sons will be (a × r) cms. above (below) the general
average height where r is the correlation coefficients between the heights of the given group of fathers and
their sons. In this case correlation is positive and since | r | ≤ 1 we have a × r ≤ a. This supports the result in
(ii) above.
But today the word regression as used in Statistics has a much wider perspective without any reference
to biometry. Regression analysis, in the general sense, means the estimation or prediction of the unknown
value of one variable from the known value of the other variable. It is one of the very important statistical
tools which is extensively used in almost all sciences — natural, social and physical. It is specially used in
business and economics to study the relationship between two or more variables that are related causally
and for estimation of demand and supply curves, cost functions, production and consumption functions, etc.
Prediction or estimation is one of the major problems in almost all spheres of human activity. The
estimation or prediction of future production, consumption, prices, investments, sales, profits, income, etc.,
are of paramount importance to a businessman or economist. Population estimates and population
projections are indispensable for efficient planning of an economy. The pharmaceutical concerns are
interested in studying or estimating the effect of new drugs on patients. Regression analysis is one of the
very scientific techniques for making such predictions. In the words of M.M. Blair “Regression analysis is
a mathematical measure of the average relationship between two or more variables in terms of the original
units of the data”.
We come across a number of inter-related events in our day-to-day life. For instance, the yield of a
crop depends on the rainfall, the cost or price of a product depends on the production and advertising
expenditure, the demand for a particular product depends on its price, expenditure of a person depends on
his income, and so on. The regression analysis confined to the study of only two variables at a time is
termed as simple regression. But quite often the values of a particular phenomenon may be affected by
multiplicity of factors. The regression analysis for studying more than two variables at a time is known as
multiple regression. However, in this chapter we shall confine ourselves to simple regression only.
9·2
BUSINESS STATISTICS
In regression analysis there are two types of variables. The variable whose value is influenced or is to
be predicted is called dependent variable and the variable which influences the values or is used for
prediction, is called independent variable. In regression analysis independent variable is also known as
regressor or predictor or explanator while the dependent variable is also known as regressed or explained
variable.
9·2. LINEAR AND NON-LINEAR REGRESSION
If the given bivariate data are plotted on a graph, the points so obtained on the scatter diagram will
more or less concentrate round a curve, called the ‘curve of regression’. Often such a curve is not distinct
and is quite confusing and sometimes complicated too. The mathematical equation of the regression curve,
usually called the regression equation, enables us to study the average change in the value of the dependent
variable for any given value of the independent variable.
If the regression curve is a straight line, we say that there is linear regression between the variables
under study. The equation of such a curve is the equation of a straight line, i.e., a first degree equation in
the variables x and y. In case of linear regression the values of the dependent variable increase by a constant
absolute amount for a unit change in the value of the independent variable. However, if the curve of
regression is not a straight line, the regression is termed as curved or non-linear regression. The regression
equation will be a functional relation between x and y involving terms in x and y of degree higher than one,
i.e., involving terms of the type x2 , y2, xy, etc. However, in this chapter we shall confine our discussion to
linear regression between two variables only.
9·3. LINES OF REGRESSION
Line of regression is the line which gives the best estimate of one variable for any given value of the
other variable. In case of two variables x and y, we shall have two lines of regression; one of y on x and the
other of x on y.
Definition. Line of regression of y on x is the line which gives the best estimate for the value of y for
any specified value of x.
Similarly, line of regression of x on y is the line which gives the best estimate for the value of x for any
specified value of y.
The term best fit is interpreted in accordance with the Principle of Least Squares which consists in
minimising the sum of the squares of the residuals or the errors of estimates, i.e., the deviations between the
given observed values of the variable and their corresponding estimated values as given by the line of best
fit. We may minimise the sum of the squares of the errors parallel to y-axis or parallel to x-axis, the former
(i.e., minimising the sum of squares of errors parallel to y-axis), gives the equation of the line of regression
of y on x and the latter, viz., minimising the sum of squares of the errors parallel to x-axis gives the equation
of the line of regression of x on y.
We shall explain below the technique of deriving the equation of the line of regression of y on x.
9·3·1. Derivation of Line of Regression of y on x. Let (x1, y 1 ), (x 2, y 2 ), …, (x n , y n ), be n pairs of
observations on the two variables x and y under study. Let
y = a + bx
… (9·1) Y
•
be the line of regression (best fit) of y on x.
•
bx
a+
For any given point P i (x i, y i) in the scatter
y=
Pi (xi, yi)
diagram, the error of estimate or residual as given by
• • •
• •
the line of best fit (9·1) is Pi Hi. Now, the x-coordinate
•
•
of Hi is same as that of Pi, viz ., x i and since Hi (xi) lies
•
on the line (9·1), the y-coordinate of Hi, i.e., H i M is
Hi (xi, a + bxi)
•
•
given by (a + bxi). Hence, the error of estimate for Pi
is given by
Pi Hi = Pi M – Hi M
O
M
X
= y i – (a + bxi)
… (9·2)
Fig. 9·1.
9·3
LINEAR REGRESSION ANALYSIS
This is the error (parallel to the y-axis) for the ith point. We will have such errors for all the points on
scatter diagram. For the points which lie above the line, the error would be positive and for the points which
lie below the line, the error would be negative.
According to the principle of least squares, we have to determine the constants a and b in (9·1) such
that the sum of the squares of the errors of estimates is minimum. In other words, we have to minimise
E =
n
n
i=1
i=1
∑ Pi Hi2 = ∑ (yi – a – bxi)2
…(9·3)
subject to variations in a and b.
We may also write E as :
E = ∑(y – ye)2 = ∑(y – a – bx)2,
…(9·3a)
where y e is the estimated value of y as given by (9·1) for given value of x and summation (∑) is taken over
the n pairs of observations.
Using the principle of maxima and minima in differential calculus, E will have an extremum
(maximum or minimum) for variations in a and b if its partial derivatives w.r.t. a and b vanish separately.
Hence from (9·3a), we get
∂E
∂E
=0
and
=0
…(9·4)
∂a
∂b
⇒
∑ y = na + b∑ x
…(9·5)
and
∑ xy = a∑ x + b∑ x2
…(9·6)
These equations are known as the normal equations for estimating a and b. The quantities ∑ x, ∑ x 2,
∑y, ∑xy can be obtained from the given set of n points (x1, y1), (x 2 , y2), …, (x n , yn) and we can solve the
equations (9·5) and (9·6) simultaneously for a and b, to get :
(∑ x2)(∑ y) – (∑ x)(∑ xy)
n∑ xy – (∑ x)(∑ y)
a=
…(9·7)
and
b=
…(9·8)
2
2
n∑ x – (∑ x)
n∑ x2 – (∑ x)2
Substituting these values of a and b from (9·7) and (9·8) in (9·1), we get the required equation of the
line of regression of y on x.
The equation of the line of regression of y on x can be obtained in a much more systematic and
–
simplified form in terms of –x, y , σx, σy and r = rxy as explained below.
Dividing both sides of (9·5) by n, the total number of pairs, we get
1
1
–
∑y = a + b . ∑x
⇒
y = a + b –x
n
n
…(9·9)
–
This implies that line of best fit, i.e., regression of y on x passes through the point ( –x , y ). Or in other
–
words, the point ( –x, y ) lies on the line of regression of y on x.
Cov (x‚ y)
From (9.8), we get :
b=
…(9·10)
σx2
We find that the equation (9·1) is in the slope-intercept form, viz., y = mx + c. Hence b represents the
slope of the line of regression of y on x. Further, we have proved in (9·9) that this line (i.e., line of
–
regression of y on x) passes through the point ( –x, y ). Hence, using the slope-point form of the equation of a
line, the required equation of the line of regression of y on x becomes :
y – –y = b (x – –x)
…(9·11)
or
But
Cov (x, y) .
y – –y =
(x – –x )
σx2
Cov (x‚ y)
r = rxy =
σxσy
…(9·12)
⇒
Cov (x, y) = r σxσy
9·4
BUSINESS STATISTICS
Substituting in (9·12), we may also write the equation of the line of regression of y on x as :
rσxσy
r σy
y – –y =
(x – –x)
⇒
y – –y =
(x – –x )
2
σx
σx
∂E
∂E
= 0 and
= 0, is only a requirement for extremum (maxima or minima) of E.
∂a
∂b
The necessary and sufficient conditions for a minima of E for variations in a and b are :
∂E
∂E
(i)
=0,
=0
∂a
∂b
…(9·13)
Remarks 1.
and
(ii)
∂2 E
>0
∂a2
and
∂2 E
∂a2
Δ=
∂2 E
∂b ∂a
∂2 E
∂a ∂b
∂2 E
∂b2
>0
…(*)
…(**)
Theorem. The solution of the least square equations (9.5) and (9.6), provides a minimum of E defined
in (9.3).
The proof of this theorem is beyond the scope of the book.
2. From (9·4) we have :
∑(y – a – bx) = 0
⇒
∑(y – ye) = 0
…(9·14)
where ye is the estimated value of y for a given value of x as given by the line of regression of y on x (9·1).
–
3. The line of regression of y on x passes through the point ( –x, y ).
4. Fitting of linear and non-linear regression (trends) is discussed in detail in Chapter 11 on ‘Time
Series Analysis’ for determining the trend values.
9·3·2. Line of Regression of x on y. The line of regression of x on y is the line which gives the best
estimate of x for any given value of y. It is also obtained by the principle of least squares on minimising the
sum of squares of the errors parallel to the x-axis (See Fig. 9·2). By starting with the equation of the form :
x = A + By,
…(9·15)
and minimising the sum of the squares of errors of estimates of x, i.e., deviations between the given values
of x and their estimates given by line of regression of x on y, viz., (9·17), i.e., minimising
E = ∑(x – A – By)2 ,
…(9·16)
we shall get the normal equations for estimating A and Y
•
B as :
•
∑ x = nA + B∑ y and ∑ xy = A∑ y + B∑ y2…(9·17)
•
•
•
Solving (9·17) simultaneously for A and B, we shall
get
•
(∑y2 )(∑x) – (∑ y)(∑ xy)
•
A=
…(9·18)
•
n∑ y2 – (∑ y)2
•
By
n∑ x y – (∑ x)(∑ y)
+
A
•
and
B=
…(9·19)
n∑ y2 – (∑ y)2
x=
Substituting these values of A and B in (9·15), we
X
shall get the required equation of line of regression of x O
Fig. 9·2.
on y.
Remark. The values of A and B obtained in (9·18) and (9·19) are same as in equations (9·7) and (9·8)
with x changed to y and y to x.
Proceeding exactly as in the case of line of regression of y on x, we shall get from (9·17) the following
results :
–x = A + B –y
(i)
…(9·20)
9·5
LINEAR REGRESSION ANALYSIS
–
This implies that the line of regression of x on y passes through the point ( –x , y ).
Cov (x‚ y) r σx
(ii)
B =
=
…(9·21)
σy2
σy
(iii) The equation of the line of regression of x on y is
x – –x = B(y – –y )
…(9·22)
Cov
(x‚
y)
⇒
x – –x =
(y – –y)
…(9·23)
σy2
rσ
⇒
x – –x = x ( y – –y )
…(9·24)
σy
The derivation of these results is left as an exercise to the reader.
Remarks 1. The regression equation (9·13) implies that the line of regression of y on x passes through
–
the point ( –x , y ). Similarly (9·24) implies that the line of regression of x on y also passes through the point
–
–
( –x , y ). Hence both the lines of regression pass through the point (–x, y ). In other words, the mean values
–
( –x , y ) can be obtained as the point of intersection of the two regression lines.
2. Why two lines of regression ? There are always two lines of regression, one of y on x and the other
of x on y. The line of regression of y on x (9·12) or (9·13) is used to estimate or predict the value of y for
any given value of x, i.e., when y is a dependent variable and x is an independent variable. The estimate so
obtained will be best in the sense that it will have the minimum possible error as defined by the principle of
least squares. We can also obtain an estimate of x for any given value of y by using equation (9·13) but the
estimate so obtained will not be best since (9·13) is obtained on minimising the sum of the squares of errors
of estimates in y and not in x. Hence to estimate or predict x for any given value of y, we use the regression
equation of x on y (9·24) which is derived on minimising the sum of the squares of errors of estimates in x.
Here x is a dependent variable and y is an independent variable. The two regression equations are not
reversible or interchangeable because of the simple reason that the basis and assumptions for deriving these
equations are quite different. The regression equation of y on x is obtained on minimising the sum of the
square of the errors parallel to the y-axis while the regression equation of x on y is obtained on minimising
the sum of squares of the errors parallel to the x-axis.
In a particular case of perfect correlation, positive or negative i.e., r = ± 1, the equation of line of
regression of y on x becomes :
σy
y – –y
–
x – –x
y – y = ± (x – –x )
⇒
=±
…(*)
σx
σy
σx
Similarly, the equation of the line of regression of x on y becomes :
y – –y
σx
–
x – –x
–
x – x = ± (y – y)
⇒
=±
,
…(**)
σy
σy
σx
which is same as (*).
Hence in case of perfect correlation, (r = ± 1), both the lines of regression coincide. Therefore, in
general, we always have two lines of regression except in the particular case of perfect correlation (r = ± 1)
when both the lines coincide and we get only one line.
9·3·3. Angle Between the Regression Lines.
If θ is the acute angle between the two lines of regression then
σxσy
1 – r2
θ = tan–1
…(9·25)
σx2 + σy2 | r |
{
In particular, if r = ± 1 then
(
(
)
(
)
)}
θ = tan–1 (0)
⇒
θ = 0 or π.
i.e., the two lines are either coincident (θ = 0) or they are parallel (θ = π). But since both the lines of
regression intersect at the point ( –x , –y ), they cannot be parallel. Hence in case of perfect correlation,
positive or negative, the two lines of regression coincide.
9·6
BUSINESS STATISTICS
If r = 0, then from (9·28),
θ = tan–1 (∞) = π/2,
i.e., if the variables are uncorrelated, the two lines of regression become perpendicular to each other.
Remarks 1. When r = 0 i.e., when x and y are uncorrelated, then the lines of regression of y on x, and x
on y are given respectively by [From (9·13) and (9·24)],
y – –y = 0
⇒
y = –y
and
x – –x = 0
⇒
x = –x
y = –y , represents a line parallel to X-axis at a distance of –y units
from the origin and x = –x , represents a line parallel to Y-axis at a
distance of –x units from the origin.
Hence, if r = 0, the two lines of regression are perpendicular to
each other and are parallel to x-axis and y-axis respectively, as
shown in Fig 9·2(a).
Y
(x, y)
y=y
•
x=x
O
X
Fig. 9·2(a ).
2. We have seen above that if r = 0 (variables uncorrelated), the two lines of regression are
perpendicular to each other and if r = ± 1, θ = 0, i.e., the two lines coincide. This leads us to the conclusion
that for higher degree of correlation between the variables, the angle between the lines is smaller, i.e., the
two lines of regression are nearer to each other. On the other hand, the angle between the lines increases,
i.e., the lines of regression move apart as the value of correlation coefficient decreases. In other words, if
the lines of regression make a larger angle, they indicate a poor degree of correlation between the variables
and ultimately for θ = π/2, i.e., the lines becoming perpendicular if no correlation exists between the
variables. Thus by plotting the lines of regression on a graph paper, we can have an approximate idea about
the degree of correlation between the two variables under study. Some illustrations are given below in
Fig. 9·3(a) to Fig. 9·3(e).
TWO LINES
COINCIDE
(r = –1)
TWO LINES
COINCIDE
(r = 1)
Y
TWO LINES
PERPENDICULAR
(r = 0)
Y
Y
TWO LINES
APART (LOW
DEGREE OF
CORRELATION)
TWO LINES
CLOSER (HIGH
DEGREE OF
CORRELATION)
Y
y=y
Y
(x, y)
x=x
O
X
Fig. 9·3(a ).
O
X O
Fig. 9·3(b).
X
Fig. 9·3(c ).
O
X
Fig. 9·3 (d).
O
X
Fig. 9·3(e ).
9·4. COEFFICIENTS OF REGRESSION
Let us consider the line of regression of y on x, viz.,
y = a + bx
The coefficient ‘b’ which is the slope of the line of regression of y on x is called the coefficient of
regression of y on x. It represents the increment in the value of the dependent variable y for a unit change
in the value of the independent variable x. In other words, it represents the rate of change of y w.r.t. x. For
notational convenience, the slope b, i.e., coefficient of regression of y on x is written as byx.
Similarly in the regression equation of x on y, viz.,
x = A + By,
the coefficient B represents the change in the value of dependent variable x for a unit change in the value of
independent variable y and is called the coefficient of regression of x on y. For notational convenience, it is
written as bxy.
Notations
byx = Coefficient of regression of y on x.
bxy = Coefficient of regression of x on y.
9·7
LINEAR REGRESSION ANALYSIS
From (9·10), the coefficient of regression of y on x is given by
Cov (x‚ y) r σy
byx =
=
[·. · Cov(x, y) = r σx σy.]
σ2
σ
x
…(9·26)
x
Similarly from (9·21), the coefficient of regression of x on y is given by :
Cov (x‚ y) r σx
bxy =
=
σy2
σy
Accordingly, the equation of the line of regression of y on x becomes
y – –y = b ( x – –x ),
yx
…(9·27)
…(9·28)
and the equation of the line of regression of x on y becomes :
x – –x = b ( y – –y )
…(9·29)
Remarks 1. For numerical computations of the equations of line of regression of y on x, and x on y, the
following formulae for the regression coefficients byx and bxy are very convenient to use.
Cov (x, y) ∑ (x – –x ) (y ––y ) n∑ xy – (∑ x )(∑ y )
byx =
=
=
…(9·30)
σx2
n∑ x2 – (∑ x)2
∑(x – –x ) 2
Cov (x‚ y) ∑ (x – –x ) (y ––y ) n∑ xy – (∑ x )(∑ y )
and
bxy =
=
=
…(9·31)
σy2
n∑y 2 – (∑y)2
∑ (y – –y ) 2
Formulae (9·30) and (9·31) are very useful for computing the values of regression coefficients from
given set of n points (x1, y1), (x2, y2), …, (xn, yn).
Other convenient formulae to be used for finding the regression coefficients for numerical problems
are :
r σy
r σx
byx =
and
bxy =
… (9·32)
σx
σy
2. Correlation coefficient between two variables x and y is a symmetrical function between x and y, i.e.,
rxy = ryx. However, the regression coefficients are not symmetric functions of x and y, i.e., byx ≠ bxy.
3. We have :
Cov (x‚ y)
Cov (x‚ y)
Cov (x‚ y)
byx =
…(*),
bxy =
…(**),
rxy =
…(***)
σx2
σy2
σ x σy
From (*) and (**), we observe that the sign of each regression coefficient byx and bxy depends on the
covariance term, since σ x > 0 and σ y > 0. If Cov (x, y) is positive, both the regression coefficients are
positive and if Cov (x, y) is negative, both the regression coefficients are negative.
4. Further, since σx > 0 and σy > 0, the sign of each of r, byx and bxy depends on the covariance term. If
Cov (x, y) is positive, all the three are positive and if Cov (x, y) is negative, all the three are negative. This
result can be stated slightly differently as follows :
The sign of correlation coefficient is same as that of the regression coefficients. If regression
coefficients are positive, r is positive and if regression coefficients are negative, r is negative.
xy
9·4·1. Theorems on Regression Coefficients
Theorem 9·1. The correlation coefficient is the geometric mean between the regression coefficients
i.e.,
r2 = byx . bxy
Cov (x‚ y)
σy
Proof. We have, byx =
=r·
2
σx
σx
Multiplying (9·34) and (9·35), we get
which establishes the result.
…(9·34)
r2 = byx . bxy
… (9·33)
and
⇒
bxy
Cov (x‚ y)
σx
=
= r·
2
σy
σy
r =±⎯
byx . bxy
√⎯⎯⎯⎯
…(9·35)
…(9·36)
9·8
BUSINESS STATISTICS
Remark. The sign to be taken before the square root is same as that of regression coefficients. If the
regression coefficients are positive, we take positive sign in (9·36) and if regression coefficients are
negative, we take negative sign in (9·36).
Theorem 9·2. If one of the regression coefficients is greater than unity (one), the other must be less
than unity.
Proof. If one of the regression coefficients is greater than 1, then the other must be less than one
because otherwise, on using (9·33), we shall get :
r2 = byx . bxy > 1,
which is impossible, since 0 ≤ r2 ≤ 1.
Theorem 9·3. The arithmetic mean of the modulus value of the regression coefficients is greater than
the modulus value of the correlation coefficient
1
2
i.e.,
[
byx
+
bxy
]>
r
…(9.37)
Theorem 9·4. Regression coefficients are independent of change of origin but not of scale.
Symbolically, if we transform from x and y to new variables u and v by change of origin and scale,
viz.,
x – a , v y – b,
=
where a, b, h (>0) and k(> 0) are constants,
…(9·38)
h
k
Then
byx = k bvu
and
bxy = h · buv
…(9·39)
h
k
In particular if we take h = k = 1, i.e., we transform the variables x and y to u and v by the relation :
u=x–a
and
v=y–b
…(9·40)
i.e., by change of origin only, then from (9·39), we get
n∑ uv – (∑ u)(∑ v)
n∑ uv – (∑ u)(∑ v)
bxy = buv =
…(9·40a)
and
byx = bvu =
…(9·40b)
n∑ v2 – (∑ v)2
n∑ u2 – (∑ u)2
These formulae are very useful for obtaining the equations of the lines of regression if the mean values
–x and / or –y come out to be in fractions or if the values of x and y are large.
Example 9·1. From the following data, obtain the two regression equations :
Sales
:
91
97
108
121
67
124
51
73
111
57
Purchases
:
71
75
69
97
70
91
39
61
80
47
Solution. Let us denote the sales by the variable X and the purchases by the variable Y.
u=
x
91
97
108
121
67
124
51
73
111
57
∑ x = 900
We have
CALCULATIONS FOR REGRESSION EQUATIONS
–
–
dy 2
dxdy
dx = x – x
dy = y – y
dx 2
1
1
1
1
1
7
5
49
25
35
18
–1
324
1
–18
31
27
961
729
837
– 23
0
529
0
0
34
21
1156
441
714
–39
–31
1521
961
1209
–17
–9
289
81
153
21
10
441
100
210
–33
–23
1089
529
759
2
2
∑ dx = 0
∑ dy = 0
∑ dx = 6360 ∑ dy = 2868 ∑ dx dy = 3900
y
71
75
69
97
70
91
39
61
80
47
∑ y = 700
–x = ∑x = 900 = 90 ;
–y = ∑y = 700 = 70
and
10
10
n
n
∑(x – –x ) (y ––y ) ∑ dx dy
byx =
=
= 3900
2
6360 = 0·6132
–
∑dx
2
∑(x – x )
9·9
LINEAR REGRESSION ANALYSIS
bxy
∑ (x – –x ) (y ––y ) ∑ dx dy
=
=
=
∑dy 2
∑(y – –y )2
3900
2868
= 1·361
Regression Equations
Equation of line of regression of y on x is
Equation of line of regression of x on y is
y – –y = b (x – –x )
x – –x = b (y – –y )
yx
⇒
⇒
⇒
xy
y – 70 = 0·6132 (x – 90)
= 0·6132x – 55·188
y = 0·6132x – 55·188 + 70·000
y = 0·6132x + 14·812
⇒
⇒
⇒
x – 90 = 1·361 (y – 70)
= 1·361y – 95·27
x = 1·361y – 95·27 + 90·00
x = 1·361y – 5·27
Remark. We have
r2 = byx bxy = 0·6132 × 1·361 = 0·8346
⇒
r = ±⎯
√⎯⎯⎯
0·8346 = ± 0·9135
But since, both the regression coefficients are positive, r must be positive. Hence, r = 0·9135.
Example 9·2. From the data given below find :
(a) The two regression coefficients.
(b) The two regression equations.
(c) The coefficient of correlation between the marks in Economics and Statistics.
(d) The most likely marks in Statistics when marks in Economics are 30.
Marks in Economics :
25
28
35
32
31
36
29
38
34
32
Marks in Statistics
:
43
46
49
41
36
32
31
30
33
39
[Himachal Pradesh Univ. M.A. (Econ.), 2003]
Solution. Let us denote the marks in Economics by the variable X and the marks in Statistics by the
variable Y.
x
y
25
28
35
32
31
36
29
38
34
32
43
46
49
41
36
32
31
30
33
39
∑ x = 320
∑ y = 380
CALCULATIONS FOR REGRESSION EQUATIONS
–
–
dx 2
dy2
dx = x – x = x – 32 dy = y – y = y – 38
–7
–4
3
0
–1
4
–3
6
2
0
∑ dx = 0
–x = ∑x = 320 = 32 ;
10
n
(a) Regression Coefficients
Here,
Coefficient of regression of y on x = byx =
5
8
11
3
–2
–6
–7
–8
–5
1
∑ dy = 0
and
49
16
9
0
1
16
9
36
4
0
∑ dx2 = 140
25
64
121
9
4
36
49
64
25
1
dxdy
–35
–32
33
0
2
–24
21
– 48
–10
0
∑ dy2 = 398 ∑ dxdy = – 93
–y = ∑y =
n
∑(x – –x ) (y – –y )
∑ dxdy – 93
=
=
–
∑ dx2 140
∑(x – x )2
380
10
= 38.
= – 0·6643
9·10
BUSINESS STATISTICS
Coefficient of regression of x on y = bxy
∑(x – –x ) (y ––y )
∑ dx dy – 93
=
=
= 398 = – 0·2337
–
∑ dy2
2
∑(y – y )
(b)
Regression Equations
Equation of the line of regression of x on y is :
Equation of the line of regression of y on x is :
–
–
x – x = bxy (y – y )
y – –y = byx (x – –x )
⇒ x – 32 = – 0·2337 (y – 38)
⇒ y – 38 = – 0·6643 (x – 32)
= – 0·2337y + 0·2337 × 38
⇒
y = – 0·6643x + 38 + 0·6643 × 32
= – 0·2337y + 8·8806
= – 0·6643x + 38 + 21·2576
⇒
x = – 0·2337y + 32 + 8·8806
⇒
y = – 0·6643x + 59·2576
…(*)
⇒
x = – 0·2337y + 40·8806
(c) Correlation Coefficient. We have
r2 = byx . bxy = (– 0·6643) × (– 0·2337) = 0·1552
⇒
r=±√
⎯⎯⎯⎯⎯
0·1552 = ± 0·394
Since both the regression coefficients are negative, r must be negative. Hence, we get r = – 0·394.
(d) In order to estimate the most likely marks in Statistics (y) when marks in Economics (x) are 30, we
shall use the line of regression of y on x viz., the equation (*). Taking x = 30 in (*), the required estimate is
given by
y = – 0·6643 × 30 + 59·2576 = –19·929 + 59·2576 = 39·3286
Hence, the most likely marks in Statistics when marks in Economics are 30, are 39·3286 ~– 39.
Example 9·3. A panel of judges A and B graded seven debators and independently awarded the
following marks :
Debator
1
2
3
4
5
6
7
Marks by A
:
40
34
28
30
44
38
31
Marks by B
:
32
39
26
30
38
34
28
An eighth debator was awarded 36 marks by Judge A while Judge B was not present.
If Judge B was also present, how many marks would you expect him to award to eighth debator
assuming same degree of relationship exists in judgement ?
[Delhi Univ. B.Com (Hons.), 1993; Himachal Pradesh Univ. M.A. (Econ.), June 1999,
Allahabad Univ. M.Com. 2002]
Solution. Let the marks awarded by Judge ‘A’ be denoted by the variable X and the marks awarded by
Judge ‘B’ by the variable Y.
CALCULATIONS FOR REGRESSION EQUATIONS
Debator
1
2
3
4
5
6
7
Total
x
40
34
28
30
44
38
31
y
32
39
26
30
38
34
28
u = x – A = x – 35
5
–1
–7
–5
9
3
–4
v = y – B = y – 30
2
9
–4
0
8
4
–2
u2
25
1
49
25
81
9
16
v2
4
81
16
0
64
16
4
uv
10
–9
28
0
72
12
8
∑u=0
∑ v = 17
∑ u2 = 206
∑ v2 = 185
∑ uv = 121
The marks awarded by Judge A to the eighth debator are given to be 36, i.e., we are given x = 36. We
want to find the marks which would have been given to the 8th debator by Judge B, if he were present. In
other words, we want to find y when x = 36. To do this we need the equation of line of regression of y on x.
In the usual notations we have :
9·11
LINEAR REGRESSION ANALYSIS
–x = A + ∑u = 35 + 0 = 35,
–y = B + ∑v = 30 + 17 = 32·4286
n
7
n
7
n∑ uv – (∑ u) (∑ v) 7 × 121 – 0 × 17 121
byx = bvu =
=
=
= 0·5874
n∑ u2 – (∑ u)2
7 × 206 – 0
206
The equation of line of regression of y on x is given by
y – –y = b (x – –x )
yx
⇒
y – 32·4286 = 0·5874 (x – 35)
= 0·5874x – 0·5874 × 35
⇒
y = 0·5874x – 20·5590 + 32·4286
⇒
y = 0·5874x + 11·8696
When x = 36,
y = 0·5874 × 36 + 11·8696 = 21·1464 + 11·8696 = 33·016
Hence, if the Judge B were also present, he would have given 33 marks to the eighth debator.
Example 9·4. A departmental store gives in-service training to its salesmen which is followed by a test.
It is considering whether it should terminate the service of any salesman who does not do well in the test.
The following data give the test scores and sales made by nine salesmen during a certain period :
Test scores
:
14
19
24
21
26
22
15
20
19
Sales (’000 Rs.) :
31
36
48
37
50
45
33
41
39
Calculate the coefficient of correlation between the test scores and the sales. Does it indicate that the
termination of services of low test scores is justified ? If the firm wants a minimum sales volume of
Rs. 30,000, what is the minimum test score that will ensure continuation of service ? Also estimate the most
probable sales volume of a salesman making a score of 28.
[Delhi Univ. B.Com. (Hons.), 2003]
Solution. Let x denote the test scores of the salesmen and y denote their corresponding sales (in ’000
Rs.)
x
14
19
24
21
26
22
15
20
19
y
31
36
48
37
50
45
33
41
39
180
360
CALCULATIONS FOR REGRESSION LINES
–
–
dx = x – x = x – 20 dy = y – y = y – 40
dx2
–6
–9
36
–1
–4
1
4
8
16
1
–3
1
6
10
36
2
5
4
–5
–7
25
0
1
0
–1
–1
1
∑ dx = 0
∑ dy = 0
∑ dx2 = 120
dy2
81
16
64
9
100
25
49
1
1
dxdy
54
04
32
– 03
60
10
35
0
01
∑ dy2 = 346
∑ dxdy = 193
–x = ∑x = 180 = 20 ;
–y = ∑y = 360 = 40
9
9
n
n
byx = Coefficient of regression of y on x
bxy = Coefficient of regression of x on y
∑ dxdy 193
∑ dxdy 193
=
=
= 1·6083
=
=
= 0·5578
∑ dx2 120
∑ dy2 346
Karl Pearson’s correlation coefficient r between x and y is given by :
Then
r2 = byx . bxy = 1·6083 × 0·5578 = 0·8971
r=±√
⎯⎯⎯⎯⎯
0·8971 = ± 0·9471
⇒
Since, the regression coefficients are positive, r is also positive.
∴
r = + 0·9471.
9·12
Aliter.
BUSINESS STATISTICS
rxy =
∑ dxdy
⎯⎯⎯⎯⎯⎯⎯⎯
√
·
∑dx 2
∑dy2
=
193
⎯
√⎯⎯⎯⎯⎯⎯
120 × 346
=
193
⎯⎯⎯⎯⎯
√
41520
193
= 203·7646
= 0·9472
Thus, we see that there is a very high degree of positive correlation between the test scores (x) and the
sales (’000 Rs.) (y). This justifies the proposal for the termination of service of those with low test scores.
Regression Equations
To obtain the test sclore (x) for given sales
To estimate the sales volume (y) of a salesman
(y), we use the equation of the line of regression of with given test score (x ), we use the line of
x on y.
regression of y on x, which is given by :
The equation of line of regression of x on y is :
y – –y = byx (x – –x )
–
–
x – x = bxy (y – y )
⇒ y – 40 = 1·6083 (x – 20)
⇒ x – 20 = 0·5578 (y – 40) = 0·5578y – 22·312
= 1·6083x – 32·1660
⇒
x = 0·5578y – 22·312 + 20
⇒
y = 1·6083x – 32·1660 + 40
⇒
x = 0·5578y – 2·312
…(*) ⇒
y = 1·6083x + 7·8340
Hence to ensure the continuation of service,
Hence the estimated sales volume of a
the minimum test score (x) corresponding to a salesman with test score of 28 is (in ’000